# The Unapologetic Mathematician

## Classes of Manifolds

As we discussed the dimension of a manifold yesterday, we passed by an interesting construction that we want to look at in more detail.

Let $(U_1,\phi_1)$ and $(U_2,\phi_2)$ be two coordinate patches on an $n$-manifold $M$. We can restrict each coordinate map to the intersection of the two patches — if the intersection is empty this is trivial, but not wrong — and get a function from one open region of $\mathbb{R}^n$ to another. Specifically, if the coordinate patches have images $V_1=\phi_1(U_1\cap U_2)$ and $V_2=\phi_2(U_1\cap U_2)$, then we get a function $\phi_1\circ\phi_2^{-1}:V_2\to V_1$. We call this the “transition function” from one coordinate patch to another, and often write it $\phi_{2,1}$

So what do we know about these transition functions? Well, $\phi_{2,1}$ is a homeomorphism from $V_2$ to $V_1$, meaning it’s continuous with a continuous inverse. But remember that $V_i\subseteq\mathbb{R}^n$, and we know a lot about real $n$-dimensional space. We know all kinds of extra structure beyond just its topology, and in particular we know what it means for a function from $\mathbb{R}^n$ to itself to be smooth.

Now, as I’ve said before, “smooth” ends up behaving like a term of art. Really what we know is what it means for a function from $\mathbb{R}^n$ to itself to be differentiable, and for that differential to be continuous, and for there to be a second differential, and so on. We even introduced classes of functions to describe this whole tower, where $C^0$ consists of continuous functions, $C^1$ consists of continuously differentiable functions, $C^2$ consists of continuously twice-differentiable functions. This pattern continuous on through $C^k$ for continuously $k$-times-differentiable functions, $C^\infty$ for functions with arbitrarily many continuous derivatives, and $C^\omega$ for “analytic” functions which are the limits of their own Taylor series.

One case we didn’t explicitly mention is $PL$: the class of “piecewise linear” functions. These are continuous functions that can be defined by breaking up their domains into a finite number of chunks and giving an affine function — a linear transformation plus a constant offset vector — on each chunk so that the borders line up continuously. These functions aren’t generally differentiable, since they turn sharp corners at the boundaries of chunks, but they’re pretty nicely behaved anyway. The class $PL$ sits between $C^0$ and $C^1$.

Anyway, for each class of functions we can define a corresponding class of manifolds. If all the transition functions between all the coordinate patches are piecewise-linear, we say that we have a “PL-manifold”. If they’re all $C^1$, we say that we have a $C^1$-manifold or a “differential manifold”. This goes up through $C^k$-manifolds, $C^\infty$-manifolds, and analytic manifolds.

In practice, “smooth” ends up meaning $C^\infty$, so that we can always take as many derivatives as we want, but really we rarely need more than a few. Still, we’ll say “smooth” and not worry about it. And from here on all our manifolds will be smooth unless explicitly stated otherwise.

February 25, 2011

## The Dimension of a Manifold

One thing that may seem clear at first blush about manifolds actually takes a little thought to be sure about. Specifically, our definition says that each point $p\in M$ is contained in some coordinate patch $U$, which comes with a coordinate map $\phi:U\to\mathbb{R}^n$. But does the $n$ here have to be the same for all points $p$?

Well, let’s start by considering what happens if two coordinate patches intersect each other. Let $U_1\subseteq M$ and $U_2\subseteq M$ be open subsets with $U_1\cap U_2$ not empty, and with coordinate maps $\phi_1:U_1\to\mathbb{R}^{n_1}$ and $\phi_2:U_2\to\mathbb{R}^{n_2}$. Now we can restrict our coordinate patches to the intersection, getting $\left(U_1\cap U_2,\phi_1\vert_{U_1\cap U_2}\right)$ and $\left(U_1\cap U_2,\phi_2\vert_{U_1\cap U_2}\right)$. These are two different coordinate patches with two different coordinate maps on the same open subset $U_1\cap U_2\subseteq M$.

So what? So now we can use $\phi_2$ in reverse to lift up from $\phi_2(U_1\cap U_2)\subseteq\mathbb{R}^{n_2}$ into $U_1\cap U_2\subseteq M$, and then use $\phi_1$ forward to drop down into $\phi_1(U_1\cap U_2)\subseteq\mathbb{R}^{n_1}$. This composition $\phi_1\circ\phi_2^{-1}$ is thus a homeomorphism from an open region in $\mathbb{R}^{n_2}$ to an open region in $\mathbb{R}^{n_1}$. And this is absolutely impossible unless $n_1=n_2$.

So now we know that any two coordinate patches that intersect must use the same value of $n$. Does this mean that we always have to use the same value of $n$? Well, not quite.

Take two distinct natural numbers $m$ and $n$. For each one, we can come up with all the coordinate patches with that dimension, and take their unions $U_m$ and $U_n$. Since the union of any collection of open sets is open, each of these sets must be open. But they can’t intersect, or else some coordinate patch with dimension $m$ and some other with dimension $n$ would have to intersect, and we just saw that they can’t.

The only way this is possible is for $U_m$ and $U_n$ to live in different connected components of $M$. And, indeed, our definition so far doesn’t rule out this possibility. We could have a two-dimensional sphere and a one-dimensional circle floating next to each other, never touching, and they would count as a manifold according to what we’ve said so far.

There are two ways around this. One is to only ever talk about connected manifolds, which automatically have a unique dimension since they only have one connected component. However, this imposes restrictions, like making it difficult to take intersections of manifolds and have the result still be a manifold, since it may suddenly be disconnected. The other alternative, which we will use, is simply to assert that all connected components of a manifold must have the same dimension.

Either way, we can specify this dimension by saying that $M$ is an $n$-dimensional manifold, or an $n$-manifold for short.

February 24, 2011

## Coordinate Patches

Let’s look back at yesterday’s example of a manifold. Not only did we cover the entire sphere by open neighborhoods with homeomorphisms to open regions of the plane, we did so in a “compatible” way, which I’ll explain soon. This notion of compatible coordinates is key to making a lot of differential topology and geometry work out right.

For the moment, though, let’s introduce a useful term. A “coordinate patch” on a manifold $M$ is an open subset $U\subseteq M$ together with a map $\phi:U\to\mathbb{R}^n$ that is a homeomorphism between $U$ and its image $\phi(U)$. Armed with this definition, we might say that a manifold is a topological space where every point can be contained in some coordinate patch. The only subtle point here is that this definition would put too much emphasis on the patches rather than on the local topology of the manifold itself.

The useful thing about a coordinate patch is that it lets us pull back coordinates from $\mathbb{R}^n$ to our manifold, or at least to the open set $U$. Let’s say $p\in U$ is sent to the point $\phi(p)\in\mathbb{R}^n$. We can now use the coordinate functions $x^i:\mathbb{R}^n\to\mathbb{R}$ to read off coordinates. In fact, when working in a particular coordinate patch, we will often abuse the notation and simply write

$\displaystyle x^i(p)=x^i(\phi(p))$

Of course, when we write $x^i(p)$ the actual number we get out for the $i$th component depends immensely on our coordinate homeomorphism $\phi$, and yet we’ve made no mention of it in our notation! This is one of the most confusing things about doing differential geometry and topology calculations involving coordinates, and it’s important to keep it in mind.

February 23, 2011

## An Example of a Manifold

Let’s be a little more explicit about our example from last time. The two-dimensional sphere consists of all the points in $\mathbb{R}^3$ of unit length. If we pick an orthonormal basis for $\mathbb{R}^3$ and write the coordinates with respect to this basis as $x$, $y$, and $z$, then we’re considering all triples $(x,y,z)$ with $x^2+y^2+z^2=1$. We want to show that this set is a manifold.

We know that we can’t hope to map the whole sphere into a plane, so we have to take some points out. Specifically, let’s remove those points with $z\leq0$, just leaving one open hemisphere. We will map this hemisphere homeomorphically to an open region in $\mathbb{R}^2$.

But this is easy: just forget the $z$-component! Sending the point $(x,y,z)$ down to the point $(x,y)$ is clearly a continuous map from the open hemisphere to the open disk with $x^2+y^2<1$. Further, for any point $(x,y)$ in the open disk, there is a unique $z\geq0$ with $x^2+y^2+z^2=1$. Indeed, we can write down

$\displaystyle(x,y)\mapsto\left(x,y,\sqrt{1-x^2-y^2}\right)$

This inverse is also continuous, and so our map is indeed a homeomorphism.

Similarly we can handle all the points in the lower hemisphere $z<0$. Again we send $(x,y,z)$ to $(x,y)$, but this time for any $(x,y)$ in the open unit disk — satisfying $x^2+y^2<0$ we can write

$\displaystyle(x,y)\mapsto\left(x,y,-\sqrt{1-x^2-y^2}\right)$

which is also continuous, so this map is again a homeomorphism.

Are we done? no, since we haven’t taken care of the points with $z=0$. But in these cases we can treat the other coordinates similarly: if $y>0$ we have our inverse pair

$\displaystyle(x,y,z)\mapsto(x,z)\qquad(x,z)\mapsto\left(x,\sqrt{1-x^2-z^2},z\right)$

while if $y<0$ we have

$\displaystyle(x,y,z)\mapsto(x,z)\qquad(x,z)\mapsto\left(x,-\sqrt{1-x^2-z^2},y,z\right)$

Similarly if $x>0$ we have

$\displaystyle(x,y,z)\mapsto(y,z)\qquad(y,z)\mapsto\left(\sqrt{1-y^2-z^2},y,z\right)$

while if $x<0$ we have

$\displaystyle(x,y,z)\mapsto(y,z)\qquad(y,z)\mapsto\left(-\sqrt{1-y^2-z^2},y,z\right)$

Now are we done? Yes, since every point on the sphere must have at least one coordinate different from zero, every point must fall into one of these six cases. Thus every point has some neighborhood which is homeomorphic to an open region in $\mathbb{R}^2$.

This same approach can be generalized to any number of dimensions. The $n$-dimensional sphere consists of those points in $\mathbb{R}^{n+1}$ with unit length. It can be covered by $2(n+1)$ open hemispheres, each with a projection just like the ones above.

February 23, 2011

## Manifolds

The central object of study in differential topology and differential geometry is a “manifold”. This is a topological space, which looks “close-up” like a real vector space. In fancier language, a manifold is “locally homeomorphic” to $\mathbb{R}^n$ for some $n$.

We know what a homeomorphism is: it’s the right notion of isomorphism in the category of topological spaces and continuous maps. More specifically, it’s a bijective function — so it has an inverse — that is continuous in both directions. Topological spaces connected by a homeomorphism are effectively “the same”.

Now, we aren’t asking that a manifold $M$ be homeomorphic to $n$-dimensional real space as a topological space. That would be far too restrictive, and we’d just get those vector spaces again, without the vector space structure of course. Instead, we want our spaces to be locally homeomorphic. That means that each point $p\in M$ should be contained in some neighborhood — an open set $N\subseteq M$ — and that this neighborhood should be homeomorphic to a neighborhood in $\mathbb{R}^n$ for some $n$.

We should note, here, that a manifold doesn’t require you to come up with explicit neighborhoods and homeomorphisms for every point. In fact, for any point $p$ there are usually many neighborhoods and many homeomorphisms that make $M$ locally homeomorphic to $\mathbb{R}^n$ at $p$, and any one of them is just as good as any other.

We should also note that many sources insist that the neighborhood be homeomorphic to all of $\mathbb{R}^n$. It turns out that this is an equivalent condition, but the connection is really more trouble than it’s worth. Given that it’s a lot easier to give explicit examples if we don’t restrict ourselves like this, we’ll just leave it out. However, to simplify language, we will often just say that a point has a neighborhood “homeomorphic to $\mathbb{R}^n$” instead of “homeomorphic to an open set in $\mathbb{R}^n$“.

As a rough example (which will be made more explicit in the future), consider the two-dimensional sphere $S^2$. This is the collection of all the points $x\in\mathbb{R}^3$ with $\lVert x\rVert=1$, with the subspace topology from $\mathbb{R}^3$. And it’s pretty well-modeled by the surface of the Earth!

So think of yourself, on the surface of the Earth (as I’m pretty sure most of you are). As you look around yourself, you can see a few miles in any direction — barring local obstacles like buildings or mountains. This gives a neighborhood, and in this neighborhood the Earth looks, well, pretty much like a flat plane! That is, locally — within a few miles of any given point — the sphere of the Earth’s surface looks like the plane $\mathbb{R}^2$. Of course we know that as we zoom out the Earth is not topologically a plane, but close up it looks like it is.

As a technical point here, the important point isn’t that the Earth looks flat, but that it looks like a plane. Currently we’re just talking about differential topology, which only concerns itself with general shapes and not with measurements of things like curvature. Even if you include bumpy features like mountains and valleys, the Earth’s surface is locally homeomorphic to a plane and that’s all that matters now.

To be a little more explicit with this sphere example, imagine a rubber ball; the surface is again a topological sphere. Now, cut a hole in the ball and stretch the hole open. If we imagine that the material of the ball is infinitely stretchy, we can pull it out until it forms a big flat disk that we can lay in the plane. Thus, for every point except the ones in that hole we’ve shown a homeomorphism from an open set containing the point to an open set in the plane. We can go back and cut a hole in a different area of the ball and do the same thing, which shows that even the points inside our original hole have a neighborhood homeomorphic to $\mathbb{R}^n$. As a side benefit, we’ve seen that most of the points on the ball have two homeomorphisms (at least!), which is perfectly fine. The definition of a manifold only asks that there be at least one.

February 22, 2011

I think that’s about enough of the representation theory of the symmetric groups for now. There’s a lot more to say, but we’ve been at this for months and I’m itching for something new.

I’m thinking next up is some differential geometry, although I’ll have to take a little side trip to cover some basic differential equations before too long. I am still looking for a good reference that gets straight to the basic existence and uniqueness theorems of differential equations. If anyone has any suggestions, I’m all ears.

February 19, 2011 Posted by | Uncategorized | 15 Comments

## More Kostka Numbers

First let’s mention a few more general results about Kostka numbers.

Among all the tableaux that partition $n$, it should be clear that $(n)\triangleright\mu$. Thus the Kostka number $K_{(n)\mu}$ is not automatically zero. In fact, I say that it’s always $1$. Indeed, the shape is a single row with $n$ entries, and the content $\mu$ gives us a list of numbers, possibly with some repeats. There’s exactly one way to arrange this list into weakly increasing order along the single row, giving $K_{(n)\mu}=1$.

On the other extreme, $\lambda\triangleright(1^n)$, so $K_{\lambda(1^n)}$ might be nonzero. The shape is given by $\lambda$, and the content $(1^n)$ gives one entry of each value from $1$ to $n$. There are no possible entries to repeat, and so any semistandard tableau with content $(1^n)$ is actually standard. Thus $K_{\lambda(1^n)}=f^\lambda$ — the number of standard tableaux of shape $\lambda$.

This means that we can decompose the module $M^{(1^n)}$:

$\displaystyle M^{(1^n)}=\bigoplus\limits_{\lambda}f^\lambda S^\lambda$

But $f^\lambda=\dim(S^\lambda)$, which means each irreducible $S_n$-module shows up here with a multiplicity equal to its dimension. That is, $M^{(1^n)}$ is always the left regular representation.

Okay, now let’s look at a full example for a single choice of $\mu$. Specifically, let $\mu=(2,2,1)$. That is, we’re looking for semistandard tableaux of various shapes, all with two entries of value $1$, two of value $2$, and one of value $3$. There are five shapes $\lambda$ with $\lambda\trianglerighteq\mu$. For each one, we will look for all the ways of filling it with the required content.

$\displaystyle\begin{array}{cccc}\lambda=(2,2,1)&\begin{array}{cc}\bullet&\bullet\\\bullet&\bullet\\\bullet&\end{array}&\begin{array}{cc}1&1\\2&2\\3&\end{array}&\\\hline\lambda=(3,1,1)&\begin{array}{ccc}\bullet&\bullet&\bullet\\\bullet&&\\\bullet&&\end{array}&\begin{array}{ccc}1&1&2\\2&&\\3&&\end{array}&\\\hline\lambda=(3,2)&\begin{array}{ccc}\bullet&\bullet&\bullet\\\bullet&\bullet&\end{array}&\begin{array}{ccc}1&1&2\\2&3&\end{array}&\begin{array}{ccc}1&1&3\\2&2&\end{array}\\\hline\lambda=(4,1)&\begin{array}{cccc}\bullet&\bullet&\bullet&\bullet\\\bullet&&&\end{array}&\begin{array}{cccc}1&1&2&2\\3&&&\end{array}&\begin{array}{cccc}1&1&2&3\\2&&&\end{array}\\\hline\lambda=(5)&\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\end{array}&\begin{array}{ccccc}1&1&2&2&3\end{array}&\end{array}$

Counting the semistandard tableaux on each row, we find the Kostka numbers. Thus we get the decomposition

$\displaystyle M^{(2,2,1)}=S^{(2,2,1)}\oplus S^{(3,1,1)}\oplus2S^{(3,2)}\oplus2S^{(4,1)}\oplus S^{(5)}$

February 18, 2011

## Kostka Numbers

Now we’ve finished our proof that the intertwinors $\bar{\theta}_T$ coming from semistandard tableauxspan the space of all intertwinors from the Specht module $S^\lambda$ to the Young tabloid module $M^\mu$. We also know that they’re linearly independent, and so they form a basis of the space of intertwinors — one for each semistandard generalized tableau.

Since the Specht modules are irreducible, we know that the dimension of this space is the multiplicity of $S^\lambda$ in $M^\mu$. And the dimension, of course, is the number of basis elements, which is the number of semistandard generalized tableaux of shape $\lambda$ and content $\mu$. This number we call the “Kostka number” $K_{\lambda\mu}$. We’ve seen that there is a decomposition

$\displaystyle M^\mu=\bigoplus\limits_{\lambda\trianglerighteq\mu}m_{\lambda\mu}S^\lambda$

Now we know that the Kostka numbers give these multiplicities, so we can write

$\displaystyle M^\mu=\bigoplus\limits_{\lambda\trianglerighteq\mu}K_{\lambda\mu}S^\lambda$

We saw before that when $\lambda=\mu$, the multiplicity is one. In terms of the Kostka numbers, this tells us that $K_{\mu\mu}=1$. Is this true? Well, the only way to fit $\mu_1$ entries with value $1$, $\mu_2$ with value $2$, and so on into a semistandard tableau of shape $\mu$ is to put all the $i$ entries on the $i$th row.

In fact, we can extend the direct sum by removing the restriction on $\lambda$:

$\displaystyle M^\mu=\bigoplus\limits_\lambda K_{\lambda\mu}S^\lambda$

This is because when $\lambda\triangleleft\mu$ we have $K_{\lambda\mu}=0$. Indeed, we must eventually have $\lambda_1+\dots+\lambda_i<\mu_1+\dots+\mu_i$, and so we can't fit all the entries with values $1$ through $i$ on the first $i$ rows of $\lambda$. We must at the very least have a repeated entry in some column, if not a descent. There are thus no semistandard generalized tableaux with shape $\lambda$ and content $\mu$ in this case.

February 17, 2011

## Intertwinors from Semistandard Tableaux Span, part 3

Now we are ready to finish our proof that the intertwinors $\bar{\theta}_T:S^\lambda\to M^\mu$ coming from semistandard generalized tableaux $T$ span the space of all intertwinors between these modules.

As usual, pick any intertwinor $\theta:S^\lambda\to M^\mu$ and write

$\displaystyle\theta(e_t)=\sum\limits_Tc_TT$

Now define the set $L_\theta$ to consist of those semistandard generalized tableaux $S$ so that $[S]\trianglelefteq[T]$ for some $T$ appearing in this sum with a nonzero coefficient. This is called the “lower order ideal” generated by the $T$ in the sum. We will prove our assertion by induction on the size of this order ideal.

If $L_\theta$ is empty, then $\theta$ must be the zero map. Indeed, our lemmas showed that if $\theta$ is not the zero map, then at least one semistandard $T$ shows up in the above sum, and this $T$ would itself belong to $L_\theta$. And of course the zero map is contained in any span.

Now, if $L_\theta$ is not empty, then there is at least some semistandard $T$ with $c_T\neq0$ in the sum. Our lemmas even show that we can pick one so that $[T]$ is maximal among all the tableaux in the sum. Let’s do that and define a new intertwinor:

$\displaystyle \theta' = \theta - c_T\bar{\theta}_T$

I say that $L_{\theta'}$ is $L_\theta$ with $T$ removed.

Every $S$ appearing in $\bar{\theta}_T(e_t)$ has $[S]\trianglelefteq[T]$, since if $T$ is semistandard then $[T]$ is the largest column equivalence class in $\theta_T(\{t\})$. Thus $L_{\theta'}$ must be a subset of $L_\theta$ since we can’t be introducing any new nonzero coefficients.

Our lemmas show that if $[S]=[T]$, then $c_S$ must appear with the same coefficient in both $\theta(e_t)$ and $c_T\bar{\theta}_T(e_t)$. That is, they must be cancelled off by the subtraction. Since $T$ is maximal there’s nothing above it that might keep it inside the ideal, and so $T\notin L_{\theta'}$.

So by induction we conclude that $\theta'$ is contained within the span of the $\bar{\theta}_T$ generated by semistandard tableaux, and thus $\theta$ must be as well.

February 14, 2011

## Intertwinors from Semistandard Tableaux Span, part 2

We continue our proof that the intertwinors $\bar{\theta}_T:S^\lambda\to M^\mu$ that come from semistandard tableaux span the space of all such intertwinors. This time I assert that if $\theta\in\hom(S^\lambda,M^\mu)$ is not the zero map, then there is some semistandard $T$ with $c_T\neq0$.

Obviously there are some nonzero coefficients; if $\theta(e_t)=0$, then

$\displaystyle\theta(e_{\pi t})=\theta(\pi e_t)=\pi\theta(e_t)=0$

which would make $\theta$ the zero map. So among the nonzero $c_T$, there are some with $[T]$ maximal in the column dominance order. I say that we can find a semistandard $T$ among them.

By the results yesterday we know that the entries in the columns of these $T$ are all distinct, so in the column tabloids we can arrange them to be strictly increasing down the columns. What we must show is that we can find one with the rows weakly increasing.

Well, let’s pick a maximal $T$ and suppose that it does have a row descent, which would keep it from being semistandard. Just like the last time we saw row descents, we get a chain of distinct elements running up the two columns:

$\displaystyle\begin{array}{ccc}a_1&\hphantom{X}&b_1\\&&\wedge\\a_2&&b_2\\&&\wedge\\\vdots&&\vdots\\&&\wedge\\a_i&>&b_i\\\wedge&&\\\vdots&&\vdots\\\wedge&&b_q\\a_p&&\end{array}$

We choose the sets $A$ and $B$ and the Garnir element $g_{A,B}$ just like before. We find

$\displaystyle g_{A,B}\left(\sum\limits_Tc_TT\right)=g_{A,B}\left(\theta(e_t)\right)=\theta\left(g_{A,B}(e_t)\right)=\theta(0)=0$

The generalized tableau $T$ must appear in $g_{A,B}(T)$ with unit coefficient, so to cancel it off there must be some other generalized tableau $T'\neq T$ with $T'=\pi T$ for some $\pi$ that shows up in $g_{A,B}$. But since this $\pi$ just interchanges some $a$ and $b$ entries, we can see that $[T']\triangleright[T]$, which contradicts the maximality of our choice of $T$.

Thus there can be no row descents in $T$, and $T$ is in fact semistandard.

February 12, 2011
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