**NSF Graduate Student Fellowship Program (GRFP).** The GRFP is a classic. This program funds master’s and doctoral degrees at institutions of higher education in the United States. The GRFP provides three years of support. Women, members of underrepresented minority groups, persons with disabilities, veterans, and undergraduate seniors are especially encouraged to apply. Their deadline is typically late October (October 22 is the 2018 deadline). Letters of reference are usually due by early November.

**National Defense Science and Engineering Graduate Fellowship (NDSEGF). **NDSEG Fellowships provide tuition, monthly living stipends, and up to \$1,000 a year in medical insurance (excluding dental and vision) for three years. NDSEG Fellowships are usable at any graduate program in the United States.

**AMS Graduate Travel Grants. **The American Mathematical Society offers partial funding for graduate students to travel to the Joint Mathematical Meetings (JMM) and the AMS sectional meetings. These grants are, of course, far smaller than those above but are useful.

**Marie Skoldowska-Curie Actions (MSCA) PhD Funding. **The MSCA funds PhD students at European institutions. The MSCA emphasizes multidisciplinary and interdisciplinary research ventures. MSCA funding lasts four years covering accommodation, travel, and living expenses. It is available to students of any nationality. The MSCA is particularly keen on supporting refugees. Applications are usually due in December.

**Ford Foundation Fellowships. **The Ford Foundation offers to graduate students in the sciences, engineering, and medicine. This fellowship provides support lasting between nine months to one year for the completion of a doctoral dissertation, by a stipend of \$25,000.

**The Rhodes Scholarship. **Contrary to common sentiment, you do not have to be the first baby born in New Haven, Connecticut to receive this funding. The Rhodes Scholarship, named after the foul imperialist Cecil B. Rhodes, funds American citizens between the ages of 18 and 25 to pursue graduate studies at Oxford University. It is not just for future politicians, people do actually study mathematics with Rhodes support; there are several at the Oxford Maths Institute right now. The Fullbright Scholarship is a similar deal used for graduate study in Australia, Finland, Iceland, Ireland, Italy, Mexico, Netherlands, South Korea, Taiwan, and the United Kingdom.

**Science, Mathematics and Research for Transformation (SMART) Scholarship. **The SMART Scholarship offers support to graduate students in engineering, mathematics, and science. It is funded by the Department of Defense. Benefits are a yearly stipend of \$25,000 to \$38,000 in addition to full tuition coverage, \$1,200 per year in health insurance, and mentoring. The SMART has a notable contingency: recipients commit to one year of civilian employment with the United States Department of Defense per academic year of funding received through the SMART Scholarship program.

**American Institute of Mathematics (AIM) Workshops. **AIM workshops take up to 28 participants. They are organized either for the purpose of conducting new research on an open problem or for the purpose of understanding an important new proof of a long-standing open problem. Workshops last one week and funding is provided for travel and room & board. There are no restrictions on citizenship.

**Note.** We do not include funding for post-doctoral fellowships. We know this is also interesting to graduate students (if not more interesting). And so, post-doctoral funding opportunities will be catalogued in a separate follow-up post, once the presently ignorant author learns about these.

**Acknowledgements. **The 2017 AMS MRC provided us with a list of some funding opportunities for graduate studies and post-doctoral work, some of which are mentioned in this post.

The idea of a scheme borrows from the definition of a manifold in differential topology. Recall that a manifold is a a topological space covered by charts diffeomorphic to open subsets of some Euclidean space $\mathbb{R}^n$. Now in a lot of ways, open subsets of $\mathbb{R}^n$ are “the place to do differential calculus” — for example, taking the derivative of a function on $\mathbb{R}^n$ at a point requires the function be defined on an open neighborhood about that point. In the same way, a *scheme* is a topological space* $X$ that locally looks like “the place to do algebra/find zeroes of functions”. However, “the place to do algebra/find zeroes of functions” is a long name, so we’ll just call this an *affine scheme* instead.

Okay, but what exactly is an affine scheme? Affine schemes are the full manifestation of a simple idea: If $A$ is a ring, we should view the elements of that ring as *functions *instead of points. If you’re thinking about rings like $\mathbb{Q}(i)$ or $\mathbb{Z}/6\mathbb{Z}$, this can be a little weird, but if you do the construction first imagining rings like $\mathbb{R}[x,y]$ it might be more clear. But the beautiful part of this is that this idea works for *any* ring — meaning that any ring can be viewed as functions on some space.

For the remainder of this article, fix a ring $A$. What space could this ring $A$ be functions on? To motivate the answer, let’s do an example from basic high school algebra. If you want to see where the polynomial $p(x) = x^3 + x^2 + x$ vanishes, your natural instinct would be to begin by factoring $p(x) = x(x^2 + x + 1)$. Then if $p(x) = 0$, you know that either $x = 0$ or $x^2 + x + 1 = 0$. Put in slightly more abstract terms, if the function $p(x)$ vanishes at some point, then either the function $x$ vanishes at that point or the function $(x^2 + x + 1)$ vanishes at that point. This is an important property we’d like our topological space $X$ to satisfy:

**Property 1: **If $f, g \in A$ and the product $fg$ vanishes at a point $x \in X$, then either $f$ vanishes at $x$ or $g$ vanishes at $x$.

This definition looks suspiciously close to the definition of a prime ideal of a ring! Recall that a *prime ideal *$\mathfrak{p}$ of a ring is an ideal where if $f,g \in A$ and $fg \in \mathfrak{p}$, then either $f \in \mathfrak{p}$ or $g \in \mathfrak{p}$. So if we assume that the “points” of our special topological space are prime ideals of the ring $\mathfrak{p}$, then we have an obvious choice for what it might mean for a function $f \in A$ to *vanish* at the point $\mathfrak{p}$ — we simply say that $f$ is defined to vanish at $\mathfrak{p}$ if $f \in \mathfrak{p}$ (or equivalently, it is zero in the ring $A/\mathfrak{p}$).

Given a ring $A$, define the *spectrum *of $A$, written $\text{Spec}(A)$, to be (as a set) the set of prime ideals of $A$. Now, I’ve promised a topology on this space, and any good topology in a subject relating to “vanishing” should have the topology related to vanishing. And luckily for us, it does! We define a set of prime ideals (think “points”) to be *closed *if it is of the form $V(S) = \{\mathfrak{q} \in \text{Spec}(A) : S \subset \mathfrak{q}\}$, for any set of “functions” $S \subset A$ (meaning, “every element of $S$ vanishes at $\mathfrak{q}$). It’s not too hard to check this forms a topology on $\text{Spec}(A)$, and is a good exercise!

This is a good definition for an affine scheme (which, recall, was a place where we could talk about where functions vanish) because in some sense the definition is “maximal” — any set of elements in the ring $A$ that can be the set of functions vanishing at a point *is* a prime ideal (since it is subject to Property 1). Moreover, functions distinguish the points in the topological space — if two points $\mathfrak{p}, \mathfrak{q}$ in the space are distinct, then there is a function that vanishes on one of the points, but not on the other one. Seeing why this is true is a good check on one’s understanding of the concept. However, you don’t necessarily get to choose *which* point — keep reading!

**Example: **Consider the ring $A = \mathbb{C}[x,y]$. What are some points in $\text{Spec}(A)$? Certainly if one chooses some $a,b \in \mathbb{C},$ then $\mathfrak{p} = (x-a, y-b)$ is a maximal (hence prime) ideal, since the quotient $A/\mathfrak{p}$ is $\mathbb{C}$, which is a field. A good way to think about the point $(x-a, y-b)$ is literally to identify it with the point $(a,b) \in \mathbb{C}^2$. Now consider the function $p(x,y) = x^4 + x^3y-xy-y^2$, and take $a = 2, b = 8$. Then $p$ vanishes at $\mathfrak{p}$ since

$p(x) = (x^3 + 2x^2 + 4x + x^2y + 2xy + 4y)(x-2) + (-x-y)(y-8)$,

which shows that $p \in \mathfrak{p}$. Note that $p$ vanishes in the normal sense at $(2, 8)$ because $p(2, 8) = 2^4 + 2^6-2^4-2^6 = 0$!

Now, are there any other points in $\text{Spec}(A)$? The answer is, yes! The polynomial $q(x,y) = x^3-y$ is an irreducible polynomial by Eisenstein’s Criterion, so the ideal $\mathfrak{q} = (x^3-y)$ is another point in $\text{Spec}(A)!$ This has an obvious interpretation in $\mathbb{C}^2$ — it’s the curve $y = x^3$! And more amazingly, notice that we can factor

$p(x,y) = x^4 + x^3y-xy-y^2 = (x^3-y)(x + y)$,

so $p \in \mathfrak{q}$. The interpretation of this is very pretty — the function $p(x,y)$ vanishes *on the entire curve* associated to $q$*. *That is very neat.

This is about half of the definition of an affine scheme. Continuing with the “rings are functions on some natural space” interpretation, one develops a sheaf on the space $\text{Spec}(A)$ so that one can, for example, talk about the function $(x^2 + y^2)(x-3)$ on the places where $x-3$ doesn’t vanish. We won’t develop this here, but this article has ideally given you a good picture of what the topological space of an affine scheme looks like, and has given you a new interpretation of rings!

*Note: It’s technically a little more than that — it’s actually a ringed space, which we won’t really get into here. Basically, every possible function you can construct on the space is built into the definition with appropriate “restriction” maps.

]]>

Once again, I was reminded that I am powerless. So the real question is, how do we do it? How do we become our own advisors? How do we free ourselves from the shackles of mathematical ignorance and the inability to help ourselves? (Okay, perhaps that last one is a bit much, but it’s a good question.) The best that I can offer is challenge by choice, a method of constantly pushing yourself just a little bit further while acknowledging, yet ultimately postponing, the goals that others are setting for you. As an example, let me explain my (continuing) progression to mathematical independence.

At the start of my first year, the goal set by my professors was clear — attempt to do everything independently. I thought this was great, but a bit out of reach. To give some personal context, I attended a small liberal arts college for my undergraduate degree, where I received a Bachelor’s of Arts in Mathematics, and then taught for three years before deciding to return to school. So even remembering that I want college ruled (not wide ruled!) notebooks was a hurdle to overcome. Hence, I set out to reach the goal of doing everything independently by first doing everything in a group. My initial goal was to read and understand every problem before our group meetings. If I could do that, then I would write ideas for how to prove the problems or maybe even write some solutions (only on scrap paper though, let’s not get too crazy). In the process, I asked a lot of questions like: how do you think about that concept? What made you try those techniques? Why are those two ideas related in your mind? At the beginning, all of these questions were unanswerable, but as the year went on we were able to discuss these processes together. By the end of the first year, my goal was stronger but still mild: write ideas for every problem and argue for valid ideas in group meetings.

During the second year, I continued to push myself. I started by trying all the problems independently and allowing the group to heavily influence modifications. By this, I mean to say that even if I came up with a different solution that we couldn’t find anything wrong with, I was still willing to change to the most commonly accepted solution. Year two ended by meeting only when necessary, handing in solutions apart from the group, following up on problems that I didn’t get correct, and trying unassigned problems. As I continue to work on independence, I am including the goal of communicating math effectively in casual conversation.

So, I offer the method of challenge by choice because you will not only meet goals set by others — you will exceed them. Eventually, you begin to set more tailored and meaningful goals. Fingers crossed, this will lead us to becoming our own advisors.

]]>I value these memories because they add texture to my life as a graduate student, providing joy or throwing it into relief.

But I’ve also had experiences that extend beyond the normal ups and downs. I listened, trying not to cry, as a professor told me that I was too slow to do theoretical work. (A year later, I won an NSF grant to do just that.) I gritted my teeth as I, the only woman in the room, was asked to sort exams into piles, while my male colleagues graded them. These experiences didn’t make me stronger, happier, more resilient, or more confident. They just wore away at my well-being.

Learning to survive graduate school as a woman in STEM—or any minority, for that matter—means finding ways to manage the effects of constant, subtle antagonism, because that antagonism won’t make you a better scientist, mathematician, or engineer.

Here are seven things that will.

**1. Reevaluate your definition of a mathematician. **Before I came to grad school, the word “mathematician” meant someone like my favorite college professor. She was creative, smart, and did good work on fascinating problems. She happened to be female, but that was beside the point—or so I thought. When I arrived at grad school, my perception of who a mathematician was* *changed. The mathematicians I interacted with were now mostly white men and I didn’t understand how significant this was until I stopped believing that I could be a mathematician. It took some reexamining of unconscious beliefs before I realized that if I wanted to become a mathematician, I’d have to start thinking of myself as one.

**2. Stop using the word “genius.”** Except to describe an expertly crafted cappuccino, I’ve all but eliminated this word from my vocabulary. It’s usually well-meant, but it reinforces troublesome perceptions of mathematics as a profession. For example: it takes a genius to do good math. Or: math comes naturally to a select few, and the rest of us shouldn’t bother. Or: if I’m not a genius, I shouldn’t do math. The last plagues women in particular, because we don’t usually think of ourselves as geniuses. Let’s start reframing the profession in terms of effort, creativity, perseverance, and intense focus—because that’s really what it’s all about.

**3. Start your own alt-boys’ club. **By which I mean: foster a supportive, engaged community that stands opposed to discrimination and celebrates one another’s achievements. Tap into your network of friends and family whose values run counter to those of the “boys’ clubs” found in so many university’s math programs: people who champion openness, diversity, celebration of achievements. Kindness. The world needs more of these communities, and in grad school, so will you. It’s difficult to be a minority in your field for a number of reasons, but the one I have struggled with the most is the isolation that comes with long hours of solitary research in a male-dominated environment. Weekly, wine-soaked dinner dates with my cousins, rock-climbing adventures with my chillest, flannel-sporting friends, and phone calls with my big sister have been my saving grace in grad school.

**4. Find an involved advisor… **Because being in grad school is hard, and being a woman in grad school is harder. A good advisor will do his or her best to share some of that burden with you, coming up with creative ways to help you succeed. My advisor had me practice my generals presentation in front of a giant, fluffy stuffed panda to shake my nerves. (Humor helps!) A good advisor will also recognize that none of us do our best work in isolation, and will do his or her best to foster a group that’s present, supportive, and actively engaged in one another’s success.

**5. … and a good therapist. **Even if it isn’t clear why so many graduate students struggle with their mental health, the fact of the matter is that they do. And even if you aren’t one of them, a therapist can help you cope with the pressures of research and exams. My therapist gave me strategies for getting through my generals, like using meditation to dispel the fight-or-flight response that kept kicking in when I practiced my presentation. She also helped me reframe stressful situations with humor and insight. Who couldn’t use more of that?

**6. Accomplish something. **Something outside of your project and coursework, that is. Research can be a long, slow grind, and a sense of progress, accomplishment, and belonging will keep you motivated. Try music. Teaching. Volunteering. Anything that gives you purpose and confidence. Particularly if you’re a minority, grad school won’t build up your confidence—a prospect even more insidious than it sounds, because it’s necessary to have confidence in order to tackle hard problems. This is, in my view, one of the most baffling paradoxes of academia: by undermining confidence and undervaluing graduate students, the academic hierarchy produces worse academics.

**7. Know why you’re in it. **When I told my advisor that I was interested in pursuing science writing and editing, it was a game-changer. I felt more comfortable talking about my career goals with my peers, and stopped feeling like an impostor. Plus, my advisor has since found writing and editing opportunities for me to hone those skills. Whatever your career goals are, communicate them—especially if they are unconventional. Mentors and friends will know which opportunities to send your way; in turn, those opportunities will keep you focused, motivated, and enthusiastic. We all come to grad school from different places and for different reasons and that *is* something worth talking about.

As a grad student, life can be hectic—with classes, homework, teaching responsibilities and more—so taking the time out of an already busy schedule to read the AMS Notices “cover to cover” is not always something that seems easily attainable. In fact, if you are like me, often when you get the email notification that the Notices is published and you quickly file it away into a folder to check out at a later date, but let’s be honest, you rarely enter that folder to peruse them after you click the button to file it away. Now as a mathematician, staying up to date is important and filing away the Notices for some future date may not be the best habit to have. That’s where this post and future posts come in handy. Every time that the Notices have a new edition available, we will be spotlighting an article that we think is of particular interest for us as graduate students. We encourage you to take a moment out of your busy schedule to read the article or, if you aren’t interested in the topic we are emphasizing, to peruse the other articles and find one that catches your eye. So without further ado, our first spotlighted article.

This month’s article is titled “What is Inquiry-Based Learning?” by Dana C. Ernst, Angie Hodge, and Stan Yoshinobu, located in the Communications section of the Notices. As we are in the throws of summer break now, we can take time to reflect on the past year of teaching, evaluating what went well (and what didn’t go as well) and all of the questions that go along with the teaching experience. In addition to reflection, we must also begin thinking about the upcoming year and how we will approach our new classes. This article is a great introduction to a classroom that looks very different from the traditional classroom experience that many students encounter, especially in math. As a brief overview, we really only need one definition to get started reading this article and that is *inquiry-based learning*. Wikipedia defines inquiry-based learning as “a form of active learning that starts by posing questions, problems, or scenarios—rather than simply presenting established facts or portraying a smooth path to knowledge.” With this definition you are now armed and ready to conquer reading this article. As you peruse this article, I challenge you to consider what ways could you incorporate some aspects of inquiry-based learning into your classroom next semester or even this summer.

To read this article and others, the Notices is available here through the AMS.

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In the summer of 2017, the AMS ran sessions entitled *Homotopy Type Theory*, *Beyond Planarity: Crossing Numbers of Graphs*, and *Dynamical Systems: Smooth, Symbolic, and Measurable* at the gorgeous Snowbird Ski Resort in Utah. I participated in the Homotopy Type Theory (HoTT) MRC. Prior to arriving, I was assigned a five-person research group and two advisors. The program began on Monday morning with opening talks and concluded with a banquet on Friday night. Over these five days we spent twenty-four total hours in intensive group work and sat in six hours of lecture. On Wednesday, as tradition dictates, we took the afternoon off to hike and explore the mountainous landscape.

After dinner on Monday we had a poster session. For this hour, some participants displayed posters describing their personal research and we bothered them with stupid questions. On Wednesday evening Dr. Emily Riehl ran an informal professional development discussion where we asked about job searches and the research process. On Thursday night, the organizers orchestrated a formal such panel and addressed questions that participants submitted by email in advance. Some questions answered: What should a research statement contain? What constitutes a good cover letter? How do you choose research problems? How do you apply for NSF grants? What are non-academic options for math PhDs? How do you balance becoming a mathematician with interests like starting a family? How do you justify your work to prospective employers? They gave detailed practical advice.

What I am most pleased with is, however, a subtler happening. Homotopy type theory is an interdisciplinary topic. The field germinated when Voevodseky, Awodey, and Warren observed that the structure of certain functional programming languages made the data types act like topological spaces [4,5]. Following a special year (2012–2013) at Princeton’s Institute for Advanced Study, it was officially known that algebraic topology could indeed be done in such a language (specifically, in a “dependent type theory”) [3]. So, both mathematicians and computer scientists attended the MRC. By Friday, everybody was on the same page despite severe background differences.

It is a pleasure to watch researchers working to overcome this language obstacle. Often, their attempts are successful—in fact, many of them are amazing. I am most excited to see the long-term collaborations ignited by this MRC and hope they will lead to a more efficient dictionary between pure mathematics and theoretical computer science.

On Friday evening, a representative from each group gave a seven minute report on the progress made and on their plans for future research. We thanked the organizers and made plans to contact each other.

A final word: Because attendees work on specific problems, the scope of an MRC session must be narrow. So, it is likely that one’s niche field will not have a session during a given summer. But, it is a good opportunity and the wise graduate student should annually check the Mathematics Research Communities’ webpage to see if a program suitable to them is offered.

**Acknowledgements **I thank the HoTT MRC organizers, J. Daniel Christensen (University of Western Ontario), Chris Kapulkin (University of Western Ontario), Daniel R. Licata (Wesleyan University), Emily Riehl (Johns Hopkins University), and Michael Shulman (University of San Diego) for making the experience worth writing about.

References

[1] Allyn Jackson, *Building a Research Career: Mathematics Research Communities, *The Notices, 2008.

[2] The American Mathematical Society, *About the Mathematics Research Communities Program, *http://www.ams.org/programs/research-communities/mrc/. Accessed 2017.

[3] *Homotopy type theory: Univalent foundations of mathematics*. The Univalent Foundations Program, Institute for Advanced Study, 2013.

[4] Steve Awodey and Michael A. Warren, *Homotopy theoretic models of identity types*, Math. Proc. Cambridge Philos. Soc. 146 (2009), no. 1, 45–55.

[5] Vladimir Voevodsky, *A very short note on homotopy λ-calculus*, notes from seminars given at Stanford University, 2006.

\[\begin{bmatrix}

1 & 2 & 3 \\

4 & 5 & 6 \\

7 & -8 & 0

\end{bmatrix}

\begin{bmatrix}

0 & 0 & 0 \\

0& 1 & 0 \\

1& 0 & 0

\end{bmatrix}

=

\begin{bmatrix}

3 & 2 & 0 \\

6 & 5 & 0 \\

0 & -8 & 0

\end{bmatrix}\]

Was there a way to have known that the first column of the product would be the third column of the first matrix?

**MATRIX TIMES VECTOR**

**Observation:** If has $n$ columns and $v$ is an -vector, then

\[Av=\begin{bmatrix}

c & c & c & \cdots \\

o & o & o & \cdots \\

l & l & l & \cdots \\

u & u & u & \cdots \\

m & m & m & \cdots \\

n & n & n & \cdots \\

\end{bmatrix}

\begin{bmatrix}

a \\

b \\

c \\

\vdots

\end{bmatrix}

=

a* \begin{bmatrix}

1^{st} \\

c \\

o \\

l \\

u \\

m \\

n

\end{bmatrix}

+b* \begin{bmatrix}

2^{nd} \\

c \\

o \\

l \\

u \\

m \\

n

\end{bmatrix}

+c* \begin{bmatrix}

3^{rd} \\

c \\

o \\

l \\

u \\

m \\

n

\end{bmatrix}+ \cdots \]

That is, the product is a linear combination of the columns with coefficients coming from the vector.

Examples:

1.

\[\begin{bmatrix}

1 & -5 & 3 \\

0 & 0.5 & 2\\

1 & -8 & 0

\end{bmatrix}

\begin{bmatrix}

2\\

-1 \\

0

\end{bmatrix}

= 2* \begin{bmatrix} 1st\\ col \end{bmatrix} + (-1)*\begin{bmatrix} 2nd\\ col \end{bmatrix} + 0 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =

2 \begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix} -\begin{bmatrix}

-5 \\

0.5 \\

-8

\end{bmatrix}+0 = \begin{bmatrix}

7 \\

-0.5 \\

10

\end{bmatrix}\]

2. What vector to multiply a matrix into, so that we get the second column of as the product?

Answer: \[ \begin{bmatrix}

0 \\

1 \\

0 \\

\vdots

\end{bmatrix}. \]

It is not only the practical calculations that benefit from this new method. Some lemmas and theorems of linear algebra become more obvious and accessible. For example:

Ax=0 has a nontrivial solution iff the columns of A are linearly dependent (as vectors).

dimension of {Ax: x ranging over all column vectors} = number of independent columns of A

Given and , has a solution iff can be written as a linear combination of columns of .

**MATRIX TIMES MATRIX**

In computing

\[AB=\begin{bmatrix}

c & c & c & \cdots \\

o & o & o & \cdots \\

l & l & l & \cdots \\

u & u & u & \cdots \\

m & m & m & \cdots \\

n & n & n & \cdots \\

\end{bmatrix}

\begin{bmatrix}

c & c & c & \cdots \\

o & o & o & \cdots \\

l & l & l & \cdots \\

u & u & u & \cdots \\

m & m & m & \cdots \\

n & n & n & \cdots \\

\end{bmatrix}

, \ \]

To get the j’th column of the product, ignore all other columns of B except the j’th, and then do A*[column j of B], using the above method. Do this for all columns and you will have the product .

Examples:

1.

\[\begin{bmatrix}

1 & 2 & 3 \\

4 & 5 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix}

1 & 0 & 1 \\

0& 1 & 1 \\

2& 3 & 0

\end{bmatrix} \]

The **first **column will be

\[ \begin{bmatrix}

1 & 2 & 3 \\

4 & 5 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix}

1 \\

0 \\

2

\end{bmatrix} =

1*\begin{bmatrix}

1 \\

4 \\

7

\end{bmatrix} + 2* \begin{bmatrix}

3 \\

-2 \\

0

\end{bmatrix} = \begin{bmatrix}

7\\

0 \\

7

\end{bmatrix} \]

The **second** column will be

\[ \begin{bmatrix}

1 & 2 & 3 \\

4 & 5 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix}

0 \\

1 \\

3

\end{bmatrix} =

1*\begin{bmatrix}

2 \\

5 \\

-4

\end{bmatrix} + 3* \begin{bmatrix}

3 \\

-2 \\

0

\end{bmatrix} = \begin{bmatrix}

11 \\

-1 \\

-4

\end{bmatrix} \]

Column 3 is computed similarly, and the full result is:

\[\begin{bmatrix}

1 & 2 & 3 \\

4 & 5 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix}

1 & 0 & 1 \\

0& 1 & 1 \\

2& 3 & 0

\end{bmatrix} = \begin{bmatrix}

7 & 11 & 3 \\

0 & -1 & 9 \\

7 & -4 & 3

\end{bmatrix} \]

Note: Most of the calculations above are doable in mind, and I have written them out to explain the procedure only.

2. Multiply \[A= \begin{bmatrix}

2 & 33 & 0 \\

0 & -1 & 9 \\

4 & -14 & 13

\end{bmatrix} \] into a matrix so that the columns 2 and 3 of A are interchanged.

Answer: \[ B=\begin{bmatrix}

1 & 0 & 0 \\

0 & 0 & 1 \\

0 & 1 & 0

\end{bmatrix} \]

3. Find the 3rd column of the product

\[\begin{bmatrix}

10 & 2 & 3 \\

4 & 0 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix}

1 & 0 & 1 \\

0& 1 & 1 \\

-2& 0 & -1

\end{bmatrix} \]

Answer:

thrid column of AB = A * third column of B. So

\[ \begin{bmatrix}

10 & 2 & 3 \\

4 & 0 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}= \begin{bmatrix} 9 \\ 6 \\ 3 \end{bmatrix} \]

(Please verify!)

4. Is the following true or false? *“If we change arrays in the second column of B, then in the product AB only the arrays in the second column may change — all other arrays on other columns will remain the same.”*

Answer: True.

Whether you are convinced that computing AB one column at a time is more practical than the conventional array-by-array multiplication or not, one cannot ignore its usefulness in helping one to “see” the linear algebra behind matrices and their multiplication. So, next time you encounter a sparse matrix save yourself tens of multiplications by zero!

**HOMEWORK**

Can you guess and prove a row-wise version of the above?

]]>

The corresponding lattices are the E8 and Leech lattices, which are important in several areas of mathematics and physics. Viazovka’s argument is simple, elegant, and canonical. The paper can be read by any person understanding only the rudiments of harmonic analysis. Stanford University’s Akshay Venkatesh has made the comment that “it seems to me much more likely than not that this function is also part of some richer story” (see the full Quanta article).

To date, optimal sphere packings are known in dimensions $2,3,8$ and $24$; with densities of $\pi/\sqrt{12}, \pi/\sqrt{18}, \pi^4/\sqrt{384}$ and $\pi^{12}/12!$, respectively. In any other dimension, it is yours for the taking!

Now, spheres are not very good packers as far as shapes go. In fact, it is conjectured that spheres are the worst (convex) packers in dimension $3$. This idea is known as Ulam’s Packing Conjecture. Is the same true in dimension $2$? No. The optimal circle packing has density $\pi/\sqrt{12} = 9068996$… and the optimal packing of regular octagons is $(4+4\sqrt{2})/(5+4\sqrt{2}) = 0.9061636$….

The problem is only interesting for convex bodies. One could take rings with arbitrarily large radii to get packings with arbitrarily small density. It is conjectured that the packing pessimum for convex disks is the regular heptagon. However, it is not even known what the optimal packings of regular heptagons is! The densest known packing is the double lattice pictured below:

This packing has density $297\left(-111+492\cos^2\left(\frac{\pi}{7}\right) – 356\cos^{22}\left(\frac{\pi}{7}\right)\right) =0.89269…$ [2]. Without knowing the optimal heptagon packing, progress on this conjecture is halted (unless, of course, the conjecture turns out to be false).

Fejes Tóth proved that for *centrally symmetric*** **convex bodies, optimal packings are always lattice packings [4]. This makes the pessimal centrally symmetric convex disk packing problem more tractable than the pessimal arbitrary convex disk problem. A body is called **centrally symmetric **if is in invariant under the transformation $x \mapsto -x.$ This is the same as possessing a 180 degree rotational symmetry. Heptagons and triangles do not exhibit this symmetry, whereas octagons and hexagons are centrally symmetric bodies.

**Conjecture (Reinhardt, 1934). **The smoothed octagon

(with its optimal packing) is a pessimum for packings of centrally symmetric convex bodies in the plane.

In March 2017, Hales posted a preprint on the arXiv “The Reinhardt Conjecture as an Optimal Control Problem” [5]. This paper lays out possible ways to resolve the Reinhardt problem by translating it into a problem in Optimal Control Theory (OTC). In what follows, we showcase this program as it presently stands.

Why did Reinhardt expect the smoothed octagon to be the worst packer? This is an unusual guess. I have never heard the smoothed octagon mentioned except as a candidate solution to the 2-D packing pessimization problem. Reinhardt himself, as I understand it, never much explained the guess. The following philosophizing was relayed to me by Hales:

Let’s start with a square. It tiles the plane with 100% density—bad guess. Maybe cutting a corner off the square will help? If we cut off its upper right corner, by central symmetry, we must also cut off its lower left corner. This is the centrally symmetric hexagon, which still tiles with 100% density. Cutting off the upper left hand corner (and therefore the lower right corner too) gives an octagon.

Moreover, solutions of the problem must be *hexagonally symmetric* [6].* *

**Definition. **Let $e_j = (\cos(2\pi j/6),\sin(2\pi j/6))$ be six equidistant points on the unit circle. Then, a disk $D \subset \mathbb{R}^2$ is **hexagonally symmetric **if it can be written as the union of six arcs $t \mapsto g(t)e_j$, $j = 0,1,…,5$ where $g : [0,t_f] \to \text{SL}_2(\mathbb{R})$ is a path in $\text{SL}_2(\mathbb{R}).$

The octagon is not hexagonally symmetric. There is only one way to round out the corners of an octagon to make it hexagonally symmetric and that is with hyperbolic arcs. This, in conjunction with failure to find a worse centrally symmetric convex packer, is our intuition for Reinhardt’s guess.

Optimal control theory is a generalization of the calculus of variations. The calculus of variations began with Johann Bernoulli’s proposal of the Brachistochrone Problem.

**Problem (Bernoulli, 1696). **Suppose that a particle of mass $m$ moves along some curve $\gamma$, in two dimensions, under the influence of gravity. Write $a,b \in \mathbb{R}^2$ for the endpoints of $\gamma$. For what $\gamma$ is the time to get from $a$ to $b$ minimized?

For a given curve $\gamma$, the time of descent is computed as $$J[y]=\int\frac{\sqrt{1+\gamma'(t)}}{\sqrt{2g\gamma(t)}}dt.$$ So, the problem is to find a function $\gamma = \gamma(t)$ which minimizes the functional $J$. Such $J$ are called **costs. **

In solving this problem, Legendre created the calculus of variations. He saw you could take a reference curve $\gamma$ and give it a slight first-order perturbation $\gamma + \epsilon \nu.$ Then, critical points of the map $\gamma \mapsto J[\gamma]$ are curves stable under small perturbations. It is a classical mathematical proof that such critical points are given by solutions to the **Euler–Lagrange equations **$$\frac{\partial L}{\partial f} – \frac{d}{dt} \frac{\partial L}{\partial \gamma’} = 0.$$

Optimal control theory addresses more general situations. Solutions to OTC problems are control functions. Control functions are prescriptions for an agent to act on the system and achieve an optimal result. For example, one can ask about the fastest way to park a car. Well, you slam on the accelerator and then slam on the break! This is not safest, but it is fastest. Of course, additional constraints can be added to the problem for safety.

An optimal control problem in finance could ask for a time-dependent control function that buys and sells stock options. In medicine, we can ask for a control function that optimally administers chemotherapeutic drugs.

The point is that the solution to the problem is a set of instructions. For car parking, it is a control function $$u(t) = \begin{cases} a_{\text{max}}, & 0 \leq t < t_0 \\ a_{\text{min}}, & t_0 \leq t \leq t_f, \end{cases}$$ where $a_{\text{max}} > 0$ denotes the maximum possible acceleration for the car, $a_{\text{min}} <0$ denotes the maximum (in absolute value) possible deceleration, $t_f$ denotes the final time and $t_0$ denotes a switching time.

The values $a_{\text{max}}$ and $a_{\text{min}}$ are determined by the makeup of the car. So, each car has a corresponding interval $[a_{\text{min}},a_{\text{max}}].$ This is called the **control set**. Our control function $u$ takes values only in extreme points of the control set. Such solutions are called **bang-bang**.

With the right eyes, smoothed octagons look like bang-bang solutions. Consider a control function for curvature, with control set $[0, \kappa_{\text{max}}].$ Smoothed polygons consist of segments of zero curvature followed abruptly by arcs of curvature $\kappa_{{max}}.$ So, smoothed polygons are described by $u$ that take values only in extreme points on $[0, \kappa_{\text{max}}].$

We have a classical OTC problem [1]:

**Dubin’s Car Problem. **Consider a car with initial position given by a vector $v \in \mathbb{R}^2.$ How does one drive the car so as to get it in some final position $w \in \mathbb{R}^2$ in minimal time?

If $v$ and $w$ are colinear, then the solution is to acceleration maximally. But, if the initial and terminal directions differ moving in a straight line is not possible. Rather the solution is to move to turn the steering wheel as hard as possible when turning and otherwise move in straight segments with maximum acceleration. In other words, we have a control set $U \subset \mathbb{R}^2$, where a point $(a,\kappa) \in U$ corresponds to acceleration at $a\text{ m/s}^2$ and turning in an arc of curvature $\kappa$. Further, the solution is bang-bang.

Restricting to a steering wheel that only turns left, there is a striking resemblance to smoothed polygons [5].

In the figure above on the left is a Dubin car trajectory and the one of the right is a hyperbolically smoothed polygon arc.

It has been shown [6] that the trajectory $g : [0,t_f] \to \text{SL}_2(\mathbb{R})$ must satisfy the equation $g=g’X$, for $X \in \mathfrak{sl}_2(\mathbb{R})$ with determinant $1$. Write $$X= \begin{pmatrix} c_{11} & c_{12} \\ c_{21} & -c_{11}. \end{pmatrix}$$ Then, if $X$ solves the Reinhardt problem $c_{21} > 0$ [6]. The set of trace $0$, determinant $1$ matrices with $c_{21}$ is the adjoint orbit of $J = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ e.g. $$X=X(x,y) = \begin{pmatrix} x/y & -x^2/y – y \\ 1/y & -x/y \end{pmatrix} = \hat{z}J\hat{z}^{-1},$$ where $\hat{z} = \begin{pmatrix} y & x \\ 0 & 1\end{pmatrix}$ [6, Lemma 2.2.1]. This gives us upper half-plane coordinates $\hat{z}J\hat{z}^{-1} \mapsto x +iy \in\mathfrak{h}$, because of the condition $1/y > 0$.

The upper-half plane is a well-known model of hyperbolic geometry. There is also a disk model $\mathbb{D} = \{w \in \mathbb{C} | |w| < 1\}$, with translation given by $w = (z-i)/(z+i) \in \mathbb{D}.$ Without loss of generality, additional constraints can be imposed on $X$: $$\sqrt{3}|c_{11}| < c_{21},\quad 3c_{12} + c_{21} < 0.$$ These are the **star inequalities**. The star region (in both the upper-half plane and the disk models) is shown below

Through the magic of Fillipov’s lemma [1], trajectories in the base manifold $\text{SL}_2(\mathbb{R}) \times \mathfrak{H}^\ast$ lift to the cotangent bundle. The problem is easier there.

Consider the star region in the upper-half plane $\mathfrak{H}^\ast.$ The path giving the smoothed octagon is following 4 edges around a triangular path centered at $i \in \mathfrak{H}^\ast$ (below, right).

The constant path at $i$ corresponds to the circle. Conjecturally, near the edges of the star region the disks approach parallelograms—computational evidence suggests this. Triangular paths with $4+3k$ edges correspond to the smoothed $(6+2k)$-gon. They all come from bang-bang controls.

This information is teased out from Pontryagin’s Maximum Principle (PMP). The PMP was discovered in 1956 by a group of Soviet mathematician’s lead by the celebrated Lev Pontryagin. It, like the Euler-Lagrange equations, is a first-order necessity condition. And it is derived in a similar way: look for trajectories for which the cost functional is stable under movements across the control set.

**Theorem (PMP). **Let $(g,z)$ be a minimizing trajectory for some measurable control $u ; [0,t_f] \to U$. Then,

- The Hamiltonian of the system vanishes along the lifted trajectory $(\lambda,u)$
- The lifted trajectory $\lambda : [0,t_f] \to T^\ast M$ is Lipschitz continuous and satisfies the ODE $$\Lambda’ = [\Lambda,X],$$ $$\nu_1′ = \frac{\partial H^+}{\partial x},$$ $$\nu_2′ = -\frac{\partial H^+}{\partial y},$$ where $H^+$ is the pointwise maximum over the control set: $$H^+(\lambda_{\text{cost}},\lambda) = \text{max}_{u \in U} H(\lambda_\text{cost},\lambda;u)$$
- Tranversality at the endpoints

This list gives us a necessity condition. Trajectories satisfying the PMP conditions are called **Pontryagin extremal**. Indeed, all smoothed $(6+2k)$-gons are Pontryagin extremals. In particular, the maximized Hamiltonian portion shows that solutions to Reinhardt’s problem must be bang-bang. To prove the Reinhardt conjecture it remains only to show there are no other extremals.

This could end in a few ways. There are higher-order generalizations of the PMP [7]. Perhaps higher-order necessity information could eliminate all other trajectories. Optimal control theory does have sufficiency tests such as Hamilton–Jacobi–Bellman’s theorem. In fact, the HJB method was used to prove global minimality for the 3-dimensional Dubin’s car problem. Yet, tragically, the Reinhardt problem does not satisfy hypothesis required to use the HJB technique. As a last resort, calculations with a supercomputer could do the job too.

My heart is with a certain topological strategy. First, by rotational symmetry of the problem, we quotient the phase space out by the group of rotations $\langle R = e^{2\pi i /3}\rangle$ around $i$. The point $i$ is the only one with isotropy, so after passing through the quotient, we remove this “orbifold singularity” to obtain an honest smooth manifold $$M = (\text{SL}_2(\mathbb{R}) \times \mathfrak{H}^\ast / \{i\}) / \langle R \rangle.$$ Because SL$_2(\mathbb{R})$ is parallelizable, the cotangent bundle projection $$\pi :((\text{SL}_2(\mathbb{R}) \times \mathfrak{sl}_2(\mathbb{R})) \times (\mathfrak{H}^\ast / \{i\} \times \mathbb{R}^2)) / \langle R \rangle \to(\text{SL}_2(\mathbb{R}) \times \mathfrak{H}^\ast / \{i\}) / \langle R \rangle$$ has contractible fibers. So, $\pi$ induces an isomorphism on homologies.

Because $\mathfrak{H}^\ast/\{i\}$ deformation retracts onto $S^1$ and $\pi_1(\text{SL}_2\mathbb{R}) = \mathbb{Z}$, we have $$H_1(T^\ast M) = \mathbb{Z} \times \mathbb{Z},$$ where $M = \text{SL}_2(\mathbb{R}) \times \mathfrak{H}^\ast/\{i\}.$ There is an evident inclusion $H_1(\mathfrak{H}^\ast/\{i\}) \to \mathbb{Z} \times \mathbb{Z}.$

The smoothed $(6k+2)$-gon lies in homology class $(1,3k+1) \in \mathbb{Z} \times \mathbb{Z}.$ Let us split by cases on the second component:

- Case $n = 3k+1, k > 0:$ The smoothed $(6+2k)$-gon lies in this class. Using Jacobi fields, one may be able to show that the smoothed polygons are pessimal in their homology class.
- Case $n \neq 1$ (mod $3$): This violates the boundary conditions of our problem. There are no extremals in these homology classes.
- Case $n = 3k + 1, k < 0:$ These are conjectured to be local maximums.
- Case $n = 1:$ This would correspond to the “homological $2$-gon.” It is conjectured these are all pathological and can be ruled out some reasons of not being a closed curve, etc.

There is the strategy. As a final remark I would like to note that if the $2$-gon exists, then it would be the global pessimum and not the smoothed octagon! Dream of homological $2$-gons.

**Acknowledgements.** Most of all I would like to thank Prof. Thomas C. Hales for teaching me everything I know about this problem. The colors images are produced by Greg Egan and the black and white images are taken from [2]. John Baez’s nlab post on this subject helped outline the first section.

References

[1] Agrachev, A.A. and Sachkov, L. *Control theory from the geometric viewpoint*, Encyclopedia of Mathematical Sciences, 87. Control Theory and Optimization, II. Springer-Verlag, Berlin (2004)

[2] Baez, J. & Egan, G. *A packing pessimization problem*, 9 2014. https://golem.ph.utexas.edu/category/2014/09/a_packing_pessimization_proble.html.

[3] Cohn, H., Kumar, A., Miller, S.D., Radchenko, D., & Viazovka, M. *The sphere packings problem in dimension 24, *https://arxiv.org/abs/1603.06518. March, 2016.

[4] Fejes Tóth, L. *On the densest packing of domains, *Proc. Kon. Ned. Akad. Wet.51 (1948), 189—192.

[5] Hales, T.C. *The Reinhardt Conjecture as a Problem in Optimal Control Theory, *https://arxiv.org/abs/1703.01352 March 3, 2017.

[6] Hales, T.C. *On the Reinhardt conjecture*, Vietnam Journal of Mathematics, 39(3):287–307, 2011. arXiv:1103.4518.

[7] Krener, A., *The high order maximal principle and its application to singular extremals,* SIAM J. Control and Opt. 15(2) (1977), pp. 256-293.

[8] Viazovska, M.S. The sphere packing problem in dimension 8, Annals of Mathematics, pages 991-1015 Volume 185 (2017), Issue 3.

]]>We then studied the specific example of re-defining the metric on the plane so that its geometry is precisely that of a 2-sphere. We saw that for measurements of angles, lengths, and areas, all we need is a dot-product on vectors. Given an open domain in the plane, once we have a dot-product, we will be able to make such measurements. Our goal in this post is to make the following definition of a manifold more tangible.

Begin with a topological space . (Note we cannot talk about any structure other than continuous maps from or to this space.) Assume that for every , there is an open neighborhood of homeomorphic to an open domain in the plane: These are called “charts”, or coordinate maps.

Given this the question now is: How do we make sense of the notions of -ness and length for a curve ? One way we might hope to do this is by using our coordinate maps. That is we say that is if is $C^1$ and we define the length of $C$ to be the length of .

As illustrated in a picture in the previous post, this definition of -ness is not a satisfactory one because some curves will lie simultaneously in two neighborhoods, say and , and there is no guarantee that if its image in is , it must also be in .

However, the two images are transformed to one another by the map (See the previous article for the reason.) Therefore, if these “transition maps” between subsets of are , then without ambiguity, we can define a subset of to be a curve if its image under any (and hence all) of the chart maps is a curve in .

The space together with the data of coordinate charts with the properties above is a 2-dimensional manifold.

As sketched above, for such manifolds, it is meaningful to talk about curves, or functions . In the latter case, we say is if is for all .

Hopefully, now, the idea is starting to make sense. A manifold is a topological space with charts whose transition maps are . For these manifolds, we can talk about second derivative of functions. A smooth manifold is one with smooth, i.e. , transition maps… Well, as is noted in [1, pg. 9], “Usually additional technical assumptions on the topological space are made to exclude pathological cases. It is customary to require that the space be Hausdorff and second countable.”

**Definition: **“A n-manifold” is a Hausdorff and second countable topological space, with charts that map into open domains of such that the transition maps are .

Where did the metric go?! How do we measure lengths?

Remember I said we might try to define the length of a curve by saying it is equal to the length of the curve lying in ? Well, this begs the question: How do we compute the length of , because might have a metric other than the usual Euclidean one. For example, recall the metric on the plane from our example in the fourth post in this series. This means we run into a problem similar to the one we had when defining -ness. If each has its own metric, then a curve on the manifold may have different lengths, depending on the chart we use for the measurement. This will make the length undefinable by looking at charts, unless, some very intricate compatibility assumptions are imposed on the metrics of the ‘s.

The good news is that one usually takes a different approach: A metric is built on the manifold upfront, rather than pieced together from collection of metrics on various ‘s that happen to magically be compatible in an complex manner. The key thing that makes this direct construction on the metric on a manifold possible, is the existence of “the tangent space”.

Fix a point , a manifold. In a chart , the point is represented by . Since is bijective and , its derivative at the point existences, and is a invertible matrix, mapping vectors centered at to vectors centered at Thus, if we fix a vector at , then in any other chart there is a corresponding vector centered at . We call this collection of vectors, one from each chart, “a tangent vector to at .” By varying in , we see that the collection of all tangent vectors to at is in one-to-one relation with the vector space of all vectors centered at , which in turn is a copy of the vector space Thus, this collection is naturally a vector space. We denote it by . The key feature is that the tangent spaces were constructed from charts and not from an ambient space in which sits in.

**Definition**: “A metric” on is a choice of an inner product on each of the tangent spaces that continuously depends on .

In our example of the plane with the metric of a sphere, attached at each , we think of a copy of the vector space . Then the inner product of the plane centered at is times the usual inner product of the plane.

What happens is that for the low dimensional tangible cases where we have a hyper-surface in a Euclidean space this abstract notion of a tangent space coincides with the “tangent plane” to the surface. For example, the tangent space to the sphere in is the copy of 2-plane touching the sphere at one point. The vectors in this tangent can be alternatively viewed as vectors in , and therefore, their inner product is defined. Therefore, by “cheating”, most visualizable spaces come with an inner product. (This is, of course, far from being the only one.) Notice we are cheating, because, we are not supposed to work with an ambient space. Here we look at the ambient space to come up with an inner product, and then forget about the ambient space again, thus, ending with a metric in the manifold sense.

**Definition: **A Riemannian manifold is a smooth manifold equipped with a smooth inner-product.

Riemannian manifolds are where we can do measurements. As said above, we deform the inner product on according to the one on the manifold: If two vectors tangent to the manifold have a dot product equal to , we define their images under to have the same inner product in the metric that will be induced on . Then, we measure lengths of curves in by this new metric. Similarly, areas can be discussed. Moreover, metric opens up the rich study of “curvature” on manifolds.

**End-note:** Why do we begin with a topological space, rather than with the sets of data I discussed earlier? That is, could we begin with a collection of open domains along with transition maps of certain smoothness, and call this collection a manifold? The answer is that, there may be many ways of patching together the pieces of data. For instance, one could always decide to leave all patches disconnected and have a union of disjoint manifolds, or decide not to patch even when the data is compatible. Thus, beginning with sets of info, there are too many possibilities, and therefore it is wise to start with a topological space as our “canvas” onto which more delicate details will be painted.

[1] Vladimir G. Ivancevic, *Applied Differential Geometry: A Modern Introduction*, (2007).

In the real numbers, we say that exactly when , that is, is *positive. *So since we’re trying to generalize the idea of “ordering”, one way is to do that is to figure out how to generalize the idea of “positive” numbers. So (and I’m being a little loose here), let’s say we only have the idea of “adding”, which we’ll denote with +, and multiplication, which we’ll denote by “*”. This is sort of what it means to be a “general field” (You might want to think about which properties of positive numbers can be defined only with +, *, or if this is too vague, just keep reading.)

Here are the two that are most important about the positive numbers–one is that if you add two positive numbers, you get a positive number again. The other is that if you multiply two positive numbers, you get a positive number again. These properties say that whatever set of positive numbers we have, it must be *closed under addition* and *closed under multiplication* resepectively.

Oh–and one other point. If we pick a generic number , we want to say that either is positive, negative, or zero. Also, has to be exactly one of them (so 10 can’t also be negative, for example). It turns out that’s all we need to make our definition of an ordered field:

**Definition: **We say that a field is an *ordered field *if it has a set (of “positive numbers”) such that:

- ( is closed under addition) If we have two elements and , then their sum is also in , that is, .
- ( is closed under multiplication) If we have two elements and , then their product is also in , that is, .
- (All nonzero numbers are positive or negative) For all in our field, exactly one of the following holds: or .

Now we’ll show something pretty cool.

**Proposition: **The complex numbers is *not *an ordered field.

Proof: To show this, we’re going to use a method called *proof by contradiction. *We’re essentially going to show that if *was *an ordered field, something bad will happen. So let’s assume *was *an ordered field and see if we can find anything weird happening.

Well one special element in that’s not in the real numbers is , where . So since , either or is positive, according to (3) above.

If was positive, then is a positive number, by (2). But again by (2), this says that is positive, so and -1 are both positive. This violates (3).

Okay, so what if was positive instead? Well, a pretty similar thing happens, since will still be positive, so we’ll get the same contradiction that 1 and -1 are both positive.

So there’s no way to order the complex numbers, at least as a field. Woah! That’s pretty neat. The mathematicians reading this may argue that if you loosen up your definition of just a set ordering, instead of a field ordering, you could put an ordering on . But instead of doing that and arguing with your computer screen, you should try to prove to yourself that any finite field can’t be ordered (as a field). It’s more fun that way.

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