## Archive for November, 2008

### Research Catchup

November 29, 2008

I’m debating about having this web space be a regular log of whatever mathematics I look at on a given day. Part of the goal will be to jot down notes on my research. I don’t know why that’s a goal, what it will accomplish, but nevertheless… Anyway, if that’s going to be the case, I thought I should get some notation going.

So, I have some understanding of $\Sigma^{\infty} C(m,V)$, the (suspension spectrum of the) space of configuration of $m$ points in a (real) vector space $V$ (with inner product). Namely, I can write it as

$\textrm{holim}_{\rho\in P} \Sigma^{\infty}(\hom(m,V)-\hom(m/\rho,V))$

Here $\hom$ just means set functions from the set $m=\{1,\ldots,m\}$ to $V$. The category $P$ that the homotopy limit is taken over is the poset of set partitions of $m$ (besides the “discrete” partition, where nothing is related to anything else, besides itself). With this in mind, the notation $\hom(m/\rho,V)$ stands for maps from $m$ to $V$ that factor through $\rho$. For shorthand, I will let $\textrm{nlc}(\rho,V)$ be the set of maps $m\to V$ that do not factor through $\rho$ (that is, $\{f|\exists x\equiv y\mod \rho\ni f(x)\neq f(y)\}$). Here “nlc” stands for “non-locally constant” (a notational choice that might make more sense later, read on). Thus, I write

$\Sigma^{\infty} C(m,V)\simeq \textrm{holim}_{\rho\in P} \Sigma^{\infty} \textrm{nlc}(\rho,V)$

I also have some understanding of $\Sigma^{\infty} \textrm{Mor}(\mathbb{R}^n,V)$, the (suspension spectrum of the) space of linear inclusions. It can be written tantalizingly similarly, as

$\textrm{holim}_{0\neq E\leq \mathbb{R}^n} \Sigma^{\infty}(\hom(\mathbb{R}^n,V)-\hom(\mathbb{R}^n/E,V)$

Here $\hom$ stands for linear maps, and we are picking out the linear maps that are non-constant on linear subspaces $E$ of $\mathbb{R}^n$. We might as well write this as $\textrm{nlc}(E,V)$, to parallel our earlier notation.

The idea of my research is to blend these two approaches and be able to understand $\textrm{Emb}(m\times \mathbb{R}^n,V)$ (well, its suspension spectrum…). I have one sort of easy way to understand it, as a product, but the understanding is not natural enough (in a strict sense). What I’m currently working on is the more natural understanding.

Given a disjoint union of affine spaces, say $A=\coprod_{i\in s} A_i$, let us say that an equivalence relation $\sim$ on $A$ is locally affine if there are vector spaces $E_i\leq V(A_i)$ ($V(A_i)$ is the vector space underlying the affine space $A_i$) such that $\sim$ factors through $A/E=\coprod_i A_i/E_i$. That is to say, if $x\sim y$ ($x\neq y$) and $x,y\in A_i$, then whenever $a,b\in A_i$ have $a-b\in \langle x-y\rangle$ (that is, the line $ab$ is parallel to the line $xy$), then $a\sim b$. We say that a locally affine partition is complete if each $E_i=V(A_i)$ (that is, the partition factors through the underlying set $s$). Thus, complete partitions are in 1-1 correspondence with set partitions of the underlying set $s$. As another bit of notation, for any equivalence relation $\rho$ on $A$, there is a unique finest coarsening of $\rho$ that is locally affine, and we denote this locally affine coarsening by $\overline{\rho}$.

Now, we make a category, $\mathscr{C}$, of complete locally affine partitions. The objects are tuples $\rho=(\rho,A_{\rho},s,\Lambda)$ where $A_{\rho}=\coprod_{i\in s}A_i$, $\Lambda$ is a set partition of the finite set $s$, and $\rho$ is the complete partition corresponding to the set partition $\Lambda$. This notation is overdetermined, of course. What you need to determine an object is… a finite set $s$ with a partition $\Lambda$ and a dimension function $d:s\rightarrow \mathbb{N}_0$. Intuitively, then, $A_i$ will be $\mathbb{R}^{d(i)}$. Maps $\rho\to \rho'$ in this category will be locally affine maps $f:A_{\rho}\to A_{\rho'}$ such that $\overline{f(\rho)}\leq \rho'$. Perhaps a word on the notation… Given any old set partition, and a function out of that set, you can take the image of the partition. This gives you a relation (not necessarily transitive) on the target set, and you then take it’s transitive closure to get an equivalence relation on the target. For us, we then take the locally affine coarsening of this image relation. The inequality $\leq$ means that $\overline{f(\rho)}$ is a coarser partition than $\rho'$. This is equivalent to saying that the underlying set partition for $\overline{f(\rho)}$ is coarser than the underlying set partition for $\rho'$.

Let $M=M_{m,n}=m\times \mathbb{R}^n=\coprod_m \mathbb{R}^n$. (So, as a reminder, my overall goal is to understand $\textrm{Emb}(M,V)$.)

Next, we get a (contravariant) functor $\textrm{nlc}(-,M)$ from this category $\mathscr{C}$ to spaces. Intuitively, given a $\rho\in \mathscr{C}$, $\textrm{nlc}(\rho,M)$ is the subspace of maps $A_{\rho}\to M$ that are affine on each component, and do not factor through the relation $\rho$ (as in our original two cases). Using this functor, we get the Grothendieck category (I think that’s the right word for it), which I denote by $\mathscr{C}_M=\mathscr{C}\ltimes \textrm{nlc}(-,M)$. Objects are pairs, $(\rho,f)$ where $\rho\in \mathscr{C}$, and $f\in \textrm{nlc}(\rho,M)$. Morphisms… are what you’d guess (being careful about the contravariance of $\textrm{nlc}$).

We also have a functor $\textrm{nlc}(-,V)$ from $\mathscr{C}$ to spaces (it’s the same thing as above, essentially). By factoring first through the projection $\mathscr{C}_M\to \mathscr{C}$ we consider $\textrm{nlc}(-,V)$ as a functor from $\mathscr{C}_M$ to spaces. It is now a standard (or so) result that $\lim_{\mathscr{C}_M}\textrm{nlc}(-,V)$ is equivalent to $\textrm{Nat}_{\mathscr{C}}(\textrm{nlc}(-,M),\textrm{nlc}(-,V))$. Of course, being after (stable) homotopy types, I’m supposed to use $\textrm{holim}$ instead of the normal limit, and end up with what my advisor calls the space of “homotopy natural transformations”. I’m also supposed to stick some $\Sigma^{\infty}$s in there, to get an equivalence with $\Sigma^{\infty} \textrm{Emb}(M,V)$.

Of course, I’m only hopeful that I will get an equivalence. I have not yet shown it’s true. In the above, some details were swept under the rug (surprise, surprise). For example, nearly all of the categories above were “topological”. Meaning, I have a space of objects, and a space of morphisms, and various maps… I’ve got a category object (pair?) in the category of spaces. And so “functor” is something you have to be a little more careful with defining, and then “holim” as well. Nevertheless… consider yourself mostly caught up. If nothing else, on the notation.

More to come? Until then, you can entertain yourself with my advisor’s work, that I am hugely indebted to. It’s on the arxiv.

### Finishing Quartics

November 18, 2008

Ok, I said I was essentially done trying to factor quartics as the composition of quadratics. The one last question I had, even if I said I didn’t really have any, was if the following statement was true:

Claim: A quartic $f=x^4+bx^3+cx^2+d$ can be factored as the composition of two quadratics iff it is symmetric around $x=-b/4$.

What I have shown so far is that if it can be factored, then it is symmetric as claimed. I’ve shown that it can be factored iff $8d=4bc-b^3$. It remains to see that symmetry implies this relation on the coefficients.

Proof: If $f$ above is symmetric about $x=-b/4$, then the function $f(x-b/4)$ is even. Plugging $x-b/4$ in to $f$, expanding everything, and gathering up terms, one sees that

$f(x-b/4)=x^4+x^2(c-\frac{b^2}{4})+x(\frac{b^3}{8}-\frac{bc}{2}+d)+(\frac{b^2c}{16}-\frac{bd}{4})$

A polynomial is even if it only has even powers of $x$. Since our $f(x-b/4)$ is even, by our symmetry assumption, the linear coefficient above must be 0. If this happens, then it is easy to rearrange the equation and see that $8d=4bc-b^3$, as desired.

So there you have it. I’d summarize this work by saying that a quartic can be written as the composition of two quadratics iff it is symmetric about a vertical line.

### Q&Q – Vertex Form

November 17, 2008

Following up yesterday’s post, I’ve spent some time with factoring quartics as the composition of two quadratics, and looking at the quadratics in vertex form to get a better graphical understanding of the problem. Also, yesterday I mentioned the apparent dis-symmetry in the freedom allowed when choosing $a_2$ as one of the free parameters, versus choosing $a_1$. I realized that came from the assumption that the leading coefficient for the quartic was positive, and probably has no further bearing on the problem.

So, let’s get back to it, as simple as possible. Our overall goal is to write $f=x^4+bx^3+cx^2+dx$ as the composition $p_1\circ p_2$ where $p_1$ and $p_2$ are quadratics. Since we now know we have enough freedom in choosing some of the coefficients, without losing any power, let’s go ahead and assume that both $p_1$ and $p_2$ are monic. I think quadratics are easier to visualize in vertex form, so write $p_i=(x-h_i)^2+k_i$, $i=1,2$ (so the vertex for $p_i$ is at $(h_i,k_i)$).

Since we’re also allowed to freely choose the constant term for $p_2$, let’s choose it to be 0. This has several wonderful implications. The constant term for $p_2$ is $h_2^2+k_2$, so if this is 0, we can set $k_2=-h_2^2$ to eliminate a parameter. Now that the constant term for $p_2$ is 0, and since we’re assuming the constant term for the composite is also 0, we see that the constant term for $p_1$ is also 0. Thus, $k_1=-h_1^2$, and we have written $p_i=(x-h_i)^2-h_i^2$. This means the vertices for the two parabolas lie on the parabola $y=-x^2$, which is somehow visually appealing. We can picture two parabolas opening up with vertices travelling around on $y=-x^2$, and make a nice movie of corresponding quartic compositions. Perhaps I’ll see what I can rig up in Maple, or something.

With these formulas for $p_1$ and $p_2$, some things are quickly apparent. It’s comforting to check that the relation $8d=4bc-b^3$ (from yesterday) is satisfied by the coefficients of the composite $p_1\circ p_2$ (here, $b,c,d$ are the coefficients of $x^3,x^2,x$ in the composite). Next, notice that

$(p_1\circ p_2)(x+h_2)=((x+h_2-h_2)^2-h_2^2-h_1)^2-h_1^2=(x^2-(h_2^2+h_1))^2-h_1^2$

which is an even function (symmetric about the $y$-axis). Since $g(x+h_2)$ is just a horizontal shift of $g(x)$ (shifting left $h_2$ units), this means that $p_1\circ p_2$ is symmetric around $x=h_2$. Unless I’ve got some implications wrong somewhere, I’m pretty sure what we’ve just shown is that a quartic can be written as the composite of two quadratics precisely when it is symmetric around a vertical axis. But I’d like to double check that statement (or let you).

Suppose you had a fixed quartic $f=x^4+bx^3+cx^2+dx$ (with $8d=4bc-b^3$) and wanted to identify the $h_1$ and $h_2$ for the quadratics that we’re talking about. It’s easy to do, with the work from yesterday. Yesterday, we said that the linear coefficient of $p_2$ was equal to $b/2$. Thus, $-2h_2=b/2$, so $h_2=-b/4$. We also found that the linear coefficient of $p_1$ would be $c-b^2/4$ . So $-2h_1=c-b^2/4$, so $h_1=\frac{b^2}{8}-\frac{c}{2}$, which is $-d/b$ if $b\neq 0$. Anyway, the point is that the vertices for the quadratics can be easily picked out from the coefficients of the quartic you started with.

Finally, a nice thing about the constant terms being 0 for these two quadratics is that then $p_i=x^2-2h_ix=x(x-2h_i)$, so one of the roots is 0, and the other is $h_i$. So the composite, $p_1\circ p_2$ will be 0 when $p_2=0$ (so $x=h_2\pm h_2$) or when $p_2=2h_1$ (so $x=h_2\pm \sqrt{h_2^2+2h_1}$). It’s also not hard to find the zeros for the first and second derivatives for $p_1\circ p_2$. The first derivative is 0 at $h_2$ and $h_2\pm \sqrt{h_2^2+h_1}$, and the second derivative is 0 at $h_2\pm \sqrt{(h_2^2+h_1)/3}$. Of course, not all of these values in square roots need be positive, but whatever. The symmetry around $x=h_2$ is readily apparent.

So, I’m feeling good about factoring quartics as quadratics. I’ve got no real questions remaining, personally. If you do, please let me know. What’s next? Well, there’s no point trying to factor a quintic as a composite, because 5 is prime, so the next logical choice would be a sixth degree polynomial as a quadratic and a cubic. As fun as that sounds, I may leave it to somebody else. I’ve got another project in mind in relation to polynomials, but you’ll just have to come back and find out about that later.

November 16, 2008

I recently wrote up a piece about factoring a given quartic polynomial as the composition of quadratics on my blogger page. I’ve been playing with it some more, and can now say things slightly better. On that page, though, I noted that we can assume that the quartic we start with is monic, with constant term 0, so I continue to assume that. I won’t go through much of the ‘proof’ of what follows, at it is covered on that blogger page.

Prop: A quartic polynomial $f=x^4+bx^3+cx^2+d$ can be factored as the composition $p_1\circ p_2$ of two quadratics iff $8d=4bc-b^3$. Moreover, when this relation on the coefficients is satisfied, there are a two-parameter family of solutions for a factorization.

Suppose the relation is satisfied, and define

$\begin{array}{rcl} p_1 &=& \frac{1}{a_2^2}x^2 +\frac{1}{a_2^2}(a_2(c-b^2/4)-2c_2)x +(\frac{c_2^2}{a_2^2}-\frac{c_2}{a_2}(c-b^2/4)) \\ p_2 &=& a_2x^2+\frac{ba_2}{2}x+c_2\end{array}$

The notation comes from assuming $p_i=a_ix^2+b_ix+c_i$, $i=1,2$. In the above, we have let $a_2$ and $c_2$ be free parameters. One could, alternatively, use $a_1$ and $c_2$ as the free parameters, obtaining polynomials

$\begin{array}{rcl}p_1 &=& a_1x^2+(\sqrt{a_1}(c-b^2/4)-2a_1c_2)x+(a_1c_2^2-c_2\sqrt{a_1}(c-b^2/4)) \\ p_2 &=& \frac{1}{\sqrt{a_1}}x^2+\frac{b}{2\sqrt{a_1}}x+c_2 \end{array}$

Clearly this requires $a_1>0$. It’s easy enough to check that the two pairs of polynomials compose to give the original quartic.

It seems like there should be something to say about the fact that the free parameter $a_1$ is more restricted (positive) than if we had chosed $a_2$ as the free parameter (it just has to be non-zero). Sadly, I don’t know what should be said.

I’ll probably continue playing with this a little. I like thinking about quadratics coming from vertex form, $y=a(x-h)^2+k$, as opposed to the standard form, used above. It’s easier to see where the quadratic is, for me, in vertex form. So I’d like to get a better visual understanding of the factorization, if I can. I smell simplifying assumption and Maple animations.

### Testing

November 15, 2008

Hello World. This is my first wordpress post. I am hoping that math is as easy as $a^2+b^2=c^2$ and that I can do things like

$e^{i\pi}+1=0.$

And I’d also like to be able to have equation arrays:

$\begin{array}{rcl} 1+1 &=& 2 \\ 2+2 & = & 4 \end{array}$

Well, that’s pretty nice. Now it’s decision time. Actually fork sumidiot.blogspot.com, and put all the math here (from now on)? Move completely over from blogger to wordpress? Just stick with blogger and delete this wp page?