# The Unapologetic Mathematician

## The Character Table as Change of Basis

Now that we’ve seen that the character table is square, we know that irreducible characters form an orthonormal basis of the space of class functions. And we also know another orthonormal basis of this space, indexed by the conjugacy classes $K\subseteq G$:

$\displaystyle\left\{\sqrt{\frac{\lvert K\rvert}{\lvert G\rvert}}f_K\right\}$

A line in the character table corresponds to an irreducible character $\chi^{(i)}$, and its entries $\chi_K^{(i)}$ tell us how to write it in terms of the basis $\{f_K\}$:

$\displaystyle\chi^{(i)}=\sum\limits_K\chi_K^{(i)}f_K$

That is, it’s a change of basis matrix from one to the other. In fact, we can modify it slightly to exploit the orthonormality as well.

When dealing with lines in the character table, we found that we can write our inner product as

$\displaystyle\langle\chi,\psi\rangle=\sum\limits_K\frac{\lvert K\rvert}{\lvert G\rvert}\overline{\chi_K}\psi_K$

So let’s modify the table to replace the entry $\chi_K^{(i)}$ with $\sqrt{\lvert K\rvert/\lvert G\rvert}\chi_K^{(i)}$. Then we have

$\displaystyle\sum\limits_K\overline{\left(\sqrt{\frac{\lvert K\rvert}{\lvert G\rvert}}\chi_K^{(i)}\right)}\left(\sqrt{\frac{\lvert K\rvert}{\lvert G\rvert}}\chi_K^{(j)}\right)=\langle\chi^{(i)},\chi^{(j)}\rangle=\delta_{i,j}$

where we make use of our orthonormality relations. That is, if we use the regular dot product on the rows of the modified character table (considered as tuples of complex numbers) we find that they’re orthonormal. But this means that the modified table is a unitary matrix, and thus its columns are orthonormal as well. We conclude that

$\displaystyle\sum\limits_i\overline{\left(\sqrt{\frac{\lvert K\rvert}{\lvert G\rvert}}\chi_K^{(i)}\right)}\left(\sqrt{\frac{\lvert K\rvert}{\lvert G\rvert}}\chi_L^{(i)}\right)=\delta_{K,L}$

where now the sum is over a set indexing the irreducible characters. We rewrite these relations as

$\displaystyle\sum\limits_i\overline{\chi_K^{(i)}}\chi_L^{(i)}=\frac{\lvert G\rvert}{\lvert K\rvert}\delta_{K,L}$

We can use these relations to help fill out character tables. For instance, let’s consider the character table of $S_3$, starting from the first two rows:

$\displaystyle\begin{array}{c|ccc}&e&(1\,2)&(1\,2\,3)\\\hline\chi^\mathrm{triv}&1&1&1\\\mathrm{sgn}&1&-1&1\\\chi^{(3)}&a&b&c\end{array}$

where we know that the third row must exist for the character table to be square. Now our new orthogonality relations tell us on the first column that

$\displaystyle1^2+1^2+a^2=6$

Since $a=\chi^{(3)}(e)$, it is a dimension, and must be positive. That is, $a=2$. On the second column we see that

$\displaystyle1^2+1^2+b^2=\frac{6}{3}=2$

and so we must have $b=0$. Finally on the third column we see that

$\displaystyle1^2+1^2+c^2=\frac{6}{2}=3$

so $c=\pm1$.

To tell the difference, we can use the new orthogonality relations on the first and third or second and third columns, or the old ones on the first and third or second and third rows. Any of them will tell us that $c=-1$, and we’ve completed the character table without worrying about constructing any representations at all.

We should take note here that the conjugacy classes index one orthonormal basis of the space of class functions, and the irreducible representations index another. Since all bases of any given vector space have the same cardinality, the set of conjugacy classes and the set of irreducible representations have the same number of elements. However, there is no reason to believe that there is any particular correspondence between the elements of the two sets. And in general there isn’t any, but we will see that in the case of symmetric groups there is a way of making just such a correspondence.

November 22, 2010

## More New Modules from Old

There are a few constructions we can make, starting with the ones from last time and applying them in certain special cases.

First off, if $V$ and $W$ are two finite-dimensional $L$-modules, then I say we can put an $L$-module structure on the space $\hom(V,W)$ of linear maps from $V$ to $W$. Indeed, we can identify $\hom(V,W)$ with $V^*\otimes W$: if $\{e_i\}$ is a basis for $V$ and $\{f_j\}$ is a basis for $W$, then we can set up the dual basis $\{\epsilon^i\}$ of $V^*$, such that $\epsilon^i(e_j)=\delta^i_j$. Then the elements $\{\epsilon^i\otimes f_j\}$ form a basis for $V^*\otimes W$, and each one can be identified with the linear map sending $e_i$ to $f_j$ and all the other basis elements of $V$ to $0$. Thus we have an inclusion $V^*\otimes W\to\hom(V,W)$, and a simple dimension-counting argument suffices to show that this is an isomorphism.

Now, since we have an action of $L$ on $V$ we get a dual action on $V^*$. And because we have actions on $V^*$ and $W$ we get one on $V^*\otimes W\cong\hom(V,W)$. What does this look like, explicitly? Well, we can write any such tensor as the sum of tensors of the form $\lambda\otimes w$ for some $\lambda\in V^*$ and $w\in W$. We calculate the action of $x\cdot(\lambda\otimes w)$ on a vector $v\in V$:

\displaystyle\begin{aligned}\left[x\cdot(\lambda\otimes w)\right](v)&=\left[(x\cdot\lambda)\otimes w\right](v)+\left[\lambda\otimes(x\cdot w)\right](v)\\&=\left[x\cdot\lambda\right](v)w+\lambda(v)(x\cdot w)\\&=-\lambda(x\cdot v)w+x\cdot(\lambda(v)w)\\&=-\left[\lambda\otimes w\right](x\cdot v)+x\cdot\left[\lambda\otimes x\right](w)\end{aligned}

In general we see that $\left[x\cdot f\right](v)=x\cdot f(v)-f(x\cdot v)$. In particular, the space of linear endomorphisms on $V$ is $\hom(V,V)$, and so it get an $L$-module structure like this.

The other case of interest is the space of bilinear forms on a module $V$. A bilinear form on $V$ is, of course, a linear functional on $V\otimes V$. And thus this space can be identified with $(V\otimes V)^*$. How does $x\in L$ act on a bilinear form $B$? Well, we can calculate:

\displaystyle\begin{aligned}\left[x\cdot B\right](v_1,v_2)&=\left[x\cdot B\right](v_1\otimes v_2)\\&=-B\left(x\cdot(v_1\otimes v_2)\right)\\&=-B\left((x\cdot v_1)\otimes v_2\right)-B\left(v_1\otimes(x\cdot v_2)\right)\\&=-B(x\cdot v_1,v_2)-B(v_1,x\cdot v_2)\end{aligned}

In particular, we can consider the case of bilinear forms on $L$ itself, where $L$ acts on itself by $\mathrm{ad}$. Here we read

$\displaystyle\left[x\cdot B\right](v_1,v_2)=-B([x,v_1],v_2)-B(v_1,[x,v_2])$

September 21, 2012

## Irreducible Modules

Sorry for the delay; it’s getting crowded around here again.

Anyway, an irreducible module for a Lie algebra $L$ is a pretty straightforward concept: it’s a module $M$ such that its only submodules are $0$ and $M$. As usual, Schur’s lemma tells us that any morphism between two irreducible modules is either $0$ or an isomorphism. And, as we’ve seen in other examples involving linear transformations, all automorphisms of an irreducible module are scalars times the identity transformation. This, of course, doesn’t depend on any choice of basis.

A one-dimensional module will always be irreducible, if it exists. And a unique — up to isomorphism, of course — one-dimensional module will always exist for simple Lie algebras. Indeed, if $L$ is simple then we know that $[L,L]=L$. Any one-dimensional representation $\phi:L\to\mathfrak{gl}(1,\mathbb{F})$ must have its image in $[\mathfrak{gl}(1,\mathbb{F}),\mathfrak{gl}(1,\mathbb{F})]=\mathfrak{sl}(1,\mathbb{F})$. But the only traceless $1\times1$ matrix is the zero matrix. Setting $\phi(x)=0$ for all $x\in L$ does indeed give a valid representation of $L$.

September 15, 2012

## Back to the Example

Let’s go back to our explicit example of $L=\mathfrak{sl}(2,\mathbb{F})$ and look at its Killing form. We first recall our usual basis:

\displaystyle\begin{aligned}x&=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\\y&=\begin{pmatrix}0&0\\1&0\end{pmatrix}\\h&=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\end{aligned}

which lets us write out matrices for the adjoint action:

\displaystyle\begin{aligned}\mathrm{ad}(x)&=\begin{pmatrix}0&0&-2\\ 0&0&0\\ 0&1&0\end{pmatrix}\\\mathrm{ad}(y)&=\begin{pmatrix}0&0&0\\ 0&0&2\\-1&0&0\end{pmatrix}\\\mathrm{ad}(h)&=\begin{pmatrix}2&0&0\\ 0&-2&0\\ 0&0&0\end{pmatrix}\end{aligned}

and from here it’s easy to calculate the Killing form. For example:

\displaystyle\begin{aligned}\kappa(x,y)&=\mathrm{Tr}\left(\mathrm{ad}(x)\mathrm{ad}(x)\right)\\&=\mathrm{Tr}\left(\begin{pmatrix}0&0&-2\\ 0&0&0\\ 0&1&0\end{pmatrix}\begin{pmatrix}0&0&0\\ 0&0&2\\-1&0&0\end{pmatrix}\right)\\&=\mathrm{Tr}\left(\begin{pmatrix}2&0&0\\ 0&0&0\\ 0&0&2\end{pmatrix}\right)\\&=4\end{aligned}

We can similarly calculate all the other values of the Killing form on basis elements.

\displaystyle\begin{aligned}\kappa(x,x)&=0\\\kappa(x,y)=\kappa(y,x)&=4\\\kappa(x,h)=\kappa(h,x)&=0\\\kappa(y,y)&=0\\\kappa(y,h)=\kappa(h,y)&=0\\\kappa(h,h)&=8\end{aligned}

So we can write down the matrix of $\kappa$:

$\displaystyle\begin{pmatrix}0&4&0\\4&0&0\\ 0&0&8\end{pmatrix}$

And we can test this for degeneracy by taking its determinant to find $-128$. Since this is nonzero, we conclude that $\kappa$ is nondegenerate, which we know means that $\mathfrak{sl}(2,\mathbb{F})$ is semisimple — at least in fields where $1+1\neq0$.

## The Killing Form

We can now define a symmetric bilinear form $\kappa$ on our Lie algebra $L$ by the formula

$\displaystyle\kappa(x,y)=\mathrm{Tr}(\mathrm{ad}(x)\mathrm{ad}(y))$

It’s symmetric because the cyclic property of the trace lets us swap $\mathrm{ad}(x)$ and $\mathrm{ad}(y)$ and get the same value. It also satisfies another identity which is referred to as “associativity”, though it may not appear like the familiar version of that property at first:

\displaystyle\begin{aligned}\kappa([x,y],z)&=\mathrm{Tr}(\mathrm{ad}([x,y])\mathrm{ad}(z))\\&=\mathrm{Tr}([\mathrm{ad}(x),\mathrm{ad}(y)]\mathrm{ad}(z))\\&=\mathrm{Tr}(\mathrm{ad}(x)[\mathrm{ad}(y),\mathrm{ad}(z)])\\&=\mathrm{Tr}(\mathrm{ad}(x)\mathrm{ad}([y,z]))\\&=\kappa(x,[y,z])\end{aligned}

Where we have used the trace identity from last time.

This is called the Killing form, named for Wilhelm Killing and not nearly so coincidentally as the Poynting vector. It will be very useful to study the structures of Lie algebras.

First, though, we want to show that the definition is well-behaved. Specifically, if $I\subseteq L$ is an ideal, then we can define $\kappa_I$ to be the Killing form of $I$. It turns out that $\kappa_I$ is just the same as $\kappa$, but restricted to take its arguments in $I$ instead of all of $L$.

A lemma: if $W\subseteq V$ is any subspace of a vector space and $\phi:V\to V$ has its image contained in $W$, then the trace of $\phi$ over $V$ is the same as its trace over $W$. Indeed, take any basis of $W$ and extend it to one of $V$; the matrix of $\phi$ with respect to this basis has zeroes for all the rows that do not correspond to the basis of $W$, so the trace may as well just be taken over $W$.

Now the fact that $I$ is an ideal means that for any $x,y\in I$ the mapping $\mathrm{ad}(x)\mathrm{ad}(y)$ is an endomorphism of $L$ sending all of $L$ into $I$. Thus its trace over $I$ is the same as its trace over all of $L$, and the Killing form on $I$ applied to $x,y\in I$ is the same as the Killing form on $L$ applied to the same two elements.

September 3, 2012 Posted by | Algebra, Lie Algebras | 5 Comments

## A Trace Criterion for Nilpotence

We’re going to need another way of identifying nilpotent endomorphisms. Let $A\subseteq B\subseteq\mathfrak{gl}(V)$ be two subspaces of endomorphisms on a finite-dimensional space $V$, and let $M$ be the collection of $x\in\mathfrak{gl}(V)$ such that $\mathrm{ad}(x)$ sends $B$ into $A$. If $x\in M$ satisfies $\mathrm{Tr}(xy)=0$ for all $y\in M$ then $x$ is nilpotent.

The first thing we do is take the Jordan-Chevalley decomposition of $x$$x=s+n$ — and fix a basis that diagonalizes $x$ with eigenvalues $a_i$. We define $E$ to be the $\mathbb{Q}$-subspace of $\mathbb{F}$ spanned by the eigenvalues. If we can prove that this space is trivial, then all the eigenvalues of $s$ must be zero, and thus $s$ itself must be zero.

We proceed by showing that any linear functional $f:E\to\mathbb{Q}$ must be zero. Taking one, we define $y\in\mathfrak{gl}(V)$ to be the endomorphism whose matrix with respect to our fixed basis is diagonal: $f(a_i)\delta_{ij}$. If $\{e_{ij}\}$ is the corresponding basis of $\mathfrak{gl}(V)$ we can calculate that

\displaystyle\begin{aligned}\left[\mathrm{ad}(s)\right](e_{ij})&=(a_i-a_j)e_{ij}\\\left[\mathrm{ad}(y)\right](e_{ij})&=(f(a_i)-f(a_j))e_{ij}\end{aligned}

Now we can find some polynomial $r(T)$ such that $r(a_i-a_j)=f(a_i)-f(a_j)$; there is no ambiguity here since if $a_i-a_j=a_k-a_l$ then the linearity of $f$ implies that

\displaystyle\begin{aligned}f(a_i)-f(a_j)&=f(a_i-a_j)\\&=f(a_k-a_l)\\&=f(a_k)-f(a_l)\end{aligned}

Further, picking $i=j$ we can see that $r(0)=0$, so $r$ has no constant term. It should be apparent that $\mathrm{ad}(y)=r\left(\mathrm{ad}(s)\right)$.

Now, we know that $\mathrm{ad}(s)$ is the semisimple part of $\mathrm{ad}(x)$, so the Jordan-Chevalley decomposition lets us write it as a polynomial in $\mathrm{ad}(x)$ with no constant term. But then we can write $\mathrm{ad}(y)=r\left(p\left(\mathrm{ad}(x)\right)\right)$. Since $\mathrm{ad}(x)$ maps $B$ into $A$, so does $\mathrm{ad}(y)$, and our hypothesis tells us that

$\displaystyle\mathrm{Tr}(xy)=\sum\limits_{i=1}^{\dim V}a_if(a_i)=0$

Hitting this with $f$ we find that the sum of the squares of the $f(a_i)$ is also zero, but since these are rational numbers they must all be zero.

Thus, as we asserted, the only possible $\mathbb{Q}$-linear functional on $E$ is zero, meaning that $E$ is trivial, all the eigenvalues of $s$ are zero, and $x$ is nipotent, as asserted.

August 31, 2012

## Uses of the Jordan-Chevalley Decomposition

Now that we’ve given the proof, we want to mention a few uses of the Jordan-Chevalley decomposition.

First, we let $A$ be any finite-dimensional $\mathbb{F}$-algebra — associative, Lie, whatever — and remember that $\mathrm{End}_\mathbb{F}(A)$ contains the Lie algebra of derivations $\mathrm{Der}(A)$. I say that if $\delta\in\mathrm{Der}(A)$ then so are its semisimple part $\sigma$ and its nilpotent part $\nu$; it’s enough to show that $\sigma$ is.

Just like we decomposed $V$ in the proof of the Jordan-Chevalley decomposition, we can break $A$ down into the eigenspaces of $\delta$ — or, equivalently, of $\sigma$. But this time we will index them by the eigenvalue: $A_a$ consists of those $x\in A$ such that $\left[\delta-aI\right]^k(x)=0$ for sufficiently large $k$.

Now we have the identity:

$\displaystyle\left[\delta-(a+b)I\right]^n(xy)=\sum\limits_{i=0}^n\binom{n}{i}\left[\delta-aI\right]^{n-i}(x)\left[\delta-bI\right]^i(y)$

which is easily verified. If a sufficiently large power of $\delta-aI$ applied to $x$ and a sufficiently large power of $\delta-bI$ applied to $y$ are both zero, then for sufficiently large $n$ one or the other factor in each term will be zero, and so the entire sum is zero. Thus we verify that $A_aA_b\subseteq A_{a+b}$.

If we take $x\in A_a$ and $y\in A_b$ then $xy\in A_{a+b}$, and thus $\sigma(xy)=(a+b)xy$. On the other hand,

\displaystyle\begin{aligned}\sigma(x)y+x\sigma(y)&=axy+bxy\\&=(a+b)xy\end{aligned}

And thus $\sigma$ satisfies the derivation property

$\displaystyle\sigma(xy)=\sigma(x)y+x\sigma(y)$

so $\sigma$ and $\nu$ are both in $\mathrm{Der}(A)$.

For the other side we note that, just as the adjoint of a nilpotent endomorphism is nilpotent, the adjoint of a semisimple endomorphism is semisimple. Indeed, if $\{v_i\}_{i=0}^n$ is a basis of $V$ such that the matrix of $x$ is diagonal with eigenvalues $\{a_i\}$, then we let $e_{ij}$ be the standard basis element of $\mathfrak{gl}(n,\mathbb{F})$, which is isomorphic to $\mathfrak{gl}(V)$ using the basis $\{v_i\}$. It’s a straightforward calculation to verify that

$\displaystyle\left[\mathrm{ad}(x)\right](e_{ij})=(a_i-a_j)e_{ij}$

and thus $\mathrm{ad}(x)$ is diagonal with respect to this basis.

So now if $x=x_s+x_n$ is the Jordan-Chevalley decomposition of $x$, then $\mathrm{ad}(x_s)$ is semisimple and $\mathrm{ad}(x_n)$ is nilpotent. They commute, since

\displaystyle\begin{aligned}\left[\mathrm{ad}(x_s),\mathrm{ad}(x_n)\right]&=\mathrm{ad}\left([x_s,x_n]\right)\\&=\mathrm{ad}(0)=0\end{aligned}

Since $\mathrm{ad}(x)=\mathrm{ad}(x_s)+\mathrm{ad}(x_n)$ is the decomposition of $\mathrm{ad}(x)$ into a semisimple and a nilpotent part which commute with each other, it is the Jordan-Chevalley decomposition of $\mathrm{ad}(x)$.

August 30, 2012

## The Jordan-Chevalley Decomposition

We recall that any linear endomorphism of a finite-dimensional vector space over an algebraically closed field can be put into Jordan normal form: we can find a basis such that its matrix is the sum of blocks that look like

$\displaystyle\begin{pmatrix}\lambda&1&&&{0}\\&\lambda&1&&\\&&\ddots&\ddots&\\&&&\lambda&1\\{0}&&&&\lambda\end{pmatrix}$

where $\lambda$ is some eigenvalue of the transformation. We want a slightly more abstract version of this, and it hinges on the idea that matrices in Jordan normal form have an obvious diagonal part, and a bunch of entries just above the diagonal. This off-diagonal part is all in the upper-triangle, so it is nilpotent; the diagonalizable part we call “semisimple”. And what makes this particular decomposition special is that the two parts commute. Indeed, the block-diagonal form means we can carry out the multiplication block-by-block, and in each block one factor is a constant multiple of the identity, which clearly commutes with everything.

More generally, we will have the Jordan-Chevalley decomposition of an endomorphism: any $x\in\mathrm{End}(V)$ can be written uniquely as the sum $x=x_s+x_n$, where $x_s$ is semisimple — diagonalizable — and $x_n$ is nilpotent, and where $x_s$ and $x_n$ commute with each other.

Further, we will find that there are polynomials $p(T)$ and $q(T)$ — each of which with no constant term — such that $p(x)=x_s$ and $q(x)=x_n$. And thus we will find that any endomorphism that commutes with $x$ with also commute with both $x_s$ and $x_n$.

Finally, if $A\subseteq B\subseteq V$ is any pair of subspaces such that $x:B\to A$ then the same is true of both $x_s$ and $x_n$.

We will prove these next time, but let’s see that this is actually true of the Jordan normal form. The first part we’ve covered.

For the second, set aside the assertion about $p$ and $q$; any endomorphism commuting with $x$ either multiplies each block by a constant or shuffles similar blocks, and both of these operations commute with both $x_n$ and $x_n$.

For the last part, we may as well assume that $B=V$, since otherwise we can just restrict to $x\vert_B\in\mathrm{End}(B)$. If $\mathrm{Im}(x)\subseteq A$ then the Jordan normal form shows us that any complementary subspace to $A$ must be spanned by blocks with eigenvalue $0$. In particular, it can only touch the last row of any such block. But none of these rows are in the range of either the diagonal or off-diagonal portions of the matrix.

August 28, 2012 Posted by | Algebra, Linear Algebra | 3 Comments

## Flags

We’d like to have matrix-oriented versions of Engel’s theorem and Lie’s theorem, and to do that we’ll need flags. I’ve actually referred to flags long, long ago, but we’d better go through them now.

In its simplest form, a flag is simply a strictly-increasing sequence of subspaces $\{V_k\}_{k=0}^n$ of a given finite-dimensional vector space. And we almost always say that a flag starts with $V_0=0$ and ends with $V_n=V$. In the middle we have some other subspaces, each one strictly including the one below it. We say that a flag is “complete” if $\dim(V_k)=k$ — and thus $n=\dim(V)$ — and for our current purposes all flags will be complete unless otherwise mentioned.

The useful thing about flags is that they’re a little more general and “geometric” than ordered bases. Indeed, given an ordered basis $\{e_k\}_{k=1}^n$ we have a flag on $V$: define $V_k$ to be the span of $\{e_i\}_{i=1}^k$. As a partial converse, given any (complete) flag we can come up with a not-at-all-unique basis: at each step let $e_k$ be the preimage in $V_k$ of some nonzero vector in the one-dimensional space $V_k/V_{k-1}$.

We say that an endomorphism of $V$ “stabilizes” a flag if it sends each $V_k$ back into itself. In fact, we saw something like this in the proof of Lie’s theorem: we build a complete flag on the subspace $W_n$, building the subspace up one basis element at a time, and then showed that each $k\in K$ stabilized that flag. More generally, we say a collection of endomorphisms stabilizes a flag if all the endomorphisms in the collection do.

So, what do Lie’s and Engel’s theorems tell us about flags? Well, Lie’s theorem tells us that if $L\subseteq\mathfrak{gl}(V)$ is solvable then it stabilizes some flag in $V$. Equivalently, there is some basis with respect to which the matrices of all elements of $L$ are upper-triangular. In other words, $L$ is isomorphic to some subalgebra of $\mathfrak{t}(\dim(V),\mathbb{F})$. We see that not only is $\mathfrak{t}(n,\mathbb{F})$ solvable, it is in a sense the archetypal solvable Lie algebra.

The proof is straightforward: Lie’s theorem tells us that $L$ has a common eigenvector $v_1\in V$. We let this span the one-dimensional subspace $V_1$ and consider the action of $L$ on the quotient $W_1=V/V_1$. Since we know that the image of $L$ in $\mathfrak{gl}(W_1)$ will again be solvable, we get a common eigenvector $w_2\in W_1$. Choosing a pre-image $v_2\in V$ with $w_2=v_2+\mathbb{F}v_1$ we get our second basis vector. We can continue like this, building up a basis of $V$ such that at each step we can write $l(v_k)\in\lambda_k(l)v_k+V_{k-1}$ for all $l\in L$ and some $\lambda_k\in L^*$.

For nilpotent $L$, the same is true — of course, nilpotent Lie algebras are automatically solvable — but Engel’s theorem tells us more: the functional $\lambda$ must be zero, and the diagonal entries of the above matrices are all zero. We conclude that any nilpotent $L$ is isomorphic to some subalgebra of $\mathfrak{n}(\dim(V),\mathbb{F})$. That is, not only is $\mathfrak{n}(n,\mathbb{F})$ nilpotent, it is the archetype of all nilpotent Lie algebras in just the same way as $\mathfrak{t}(n,\mathbb{F})$ is the archetypal solvable Lie algebra.

More generally, if $L$ is any solvable (nilpotent) Lie algebra and $\phi:L\to\mathfrak{gl}(V)$ is any finite-dimensional representation of $L$, then we know that the image $\phi(L)$ is a solvable (nilpotent) linear Lie algebra acting on $V$, and thus it must stabilize some flag of $V$. As a particular example, consider the adjoint action $\mathrm{ad}:L\to\mathfrak{gl}(L)$; a subspace of $L$ invariant under the adjoint action of $L$ is just the same thing as an ideal of $L$, so we find that there must be some chain of ideals:

$\displaystyle 0=I_0\subseteq I_1\subseteq\dots\subseteq I_{n-1}\subseteq I_n=L$

where $\dim(I_k)=k$. Given such a chain, we can of course find a basis of $L$ with respect to which the matrices of the adjoint action are all in $\mathfrak{t}(\dim(L),\mathbb{F})$ ($\mathfrak{n}(\dim(L),\mathbb{F})$).

In either case, we find that $[L,L]$ is nilpotent. Indeed, if $L$ is already nilpotent this is trivial. But if $L$ is merely solvable, we see that the matrices of the commutators $[\mathrm{ad}(x),\mathrm{ad}(y)]$ for $x,y\in L$ lie in

$\displaystyle [\mathfrak{t}(\dim(L),\mathbb{F}),\mathfrak{t}(\dim(L),\mathbb{F})]=\mathfrak{n}(\dim(L),\mathbb{F})$

But since $\mathrm{ad}$ is a homomorphism, this is the matrix of $\mathrm{ad}([x,y])$ acting on $L$, and obviously its action on the subalgebra $[L,L]$ is nilpotent as well. Thus each element of $[L,L]$ is ad-nilpotent, and Engel’s theorem then tells us that $[L,L]$ is a nilpotent Lie algebra.

## Lie’s Theorem

The lemma leading to Engel’s theorem boils down to the assertion that there is some common eigenvector for all the endomorphisms in a nilpotent linear Lie algebra $L\subseteq\mathfrak{gl}(V)$ on a finite-dimensional nonzero vector space $V$. Lie’s theorem says that the same is true of solvable linear Lie algebras. Of course, in the nilpotent case the only possible eigenvalue was zero, so we may find things a little more complicated now. We will, however, have to assume that $\mathbb{F}$ is algebraically closed and that no multiple of the unit in $\mathbb{F}$ is zero.

We will proceed by induction on the dimension of $L$ using the same four basic steps as in the lemma: find an ideal $K\subseteq L$ of codimension one, so we can write $L=K+\mathbb{F}z$ for some $z\in K\setminus L$; find common eigenvectors for $K$; find a subspace of such common eigenvectors stabilized by $L$; find in that space an eigenvector for $z$.

First, solvability says that $L$ properly includes $[L,L]$, or else the derived series wouldn’t be able to even start heading towards $0$. The quotient $L/[L,L]$ must be abelian, with all brackets zero, so we can pick any subspace of this quotient with codimension one and it will be an ideal. The preimage of this subspace under the quotient projection will then be an ideal $K\subseteq L$ of codimension one.

Now, $K$ is a subalgebra of $L$, so we know it’s also solvable, so induction tells us that there’s a common eigenvector $v\in V$ for the action of $K$. If $K$ is zero, then $L$ must be one-dimensional abelian, in which case the proof is obvious. Otherwise there is some linear functional $\lambda\in K^*$ defined by

$\displaystyle k(v)=\lambda(k)v$

Of course, $v$ is not the only such eigenvector; we define the (nonzero) subspace $W$ by

$\displaystyle W=\{w\in V\vert\forall k\in K, k(w)=\lambda(k)w\}$

Next we must show that $L$ sends $W$ back into itself. To see this, pick $l\in L$ and $k\in K$ and check that

\displaystyle\begin{aligned}k(l(w))&=l(k(w))-[l,k](w)\\&=l(\lambda(k)w)-\lambda([l,k])w\\&=\lambda(k)l(w)-\lambda([l,k])w\end{aligned}

But if $l(w)\in W$, then we’d have $k(l(w))=\lambda(k)l(w)$; we need to verify that $\lambda([l,k])=0$. In the nilpotent case — Engel’s theorem — the functional $\lambda$ was constantly zero, so this was easy, but it’s a bit harder here.

Fixing $w\in W$ and $l\in L$ we pick $n$ to be the first index where the collection $\{l^i(w)\}_{i=0}^n$ is linearly independent — the first one where we can express $l^n(w)$ as the linear combination of all the previous $l^i(w)$. If we write $W_i$ for the subspace spanned by the first $i$ of these vectors, then the dimension of $W_i$ grows one-by-one until we get to $\dim(W_n)=n$, and $W_{n+i}=W_n$ from then on.

I say that each of the $W_i$ are invariant under each $k\in K$. Indeed, we can prove the congruence

$\displaystyle k(l^i(w))\equiv\lambda(k)l^i(w)\quad\mod W_i$

that is, $k$ acts on $l^i(w)$ by multiplication by $\lambda(k)$, plus some “lower-order terms”. For $i=0$ this is the definition of $\lambda$; in general we have

\displaystyle\begin{aligned}k(l^i(w))&=k(l(l^{i-1}(w)))\\&=l(k(l^{i-1}(w)))-[l,k](l^{i-1}(w))\\&=\lambda(k)l^i(w)+l(w')-\lambda([l,k])l^{i-1}(w)-w''\end{aligned}

for some $w',w''\in W_{i-1}$.

And so we conclude that, using the obvious basis of $W_n$ the action of $k$ on this subspace is in the form of an upper-triangular matrix with $\lambda(k)$ down the diagonal. The trace of this matrix is $n\lambda(k)$. And in particular, the trace of the action of $[l,k]$ on $W_n$ is $n\lambda([l,k])$. But $l$ and $k$ both act as endomorphisms of $W_n$ — the one by design and the other by the above proof — and the trace of any commutator is zero! Since $n$ must have an inverse we conclude that $\lambda([l,k])=0$.

Okay so that checks out that the action of $L$ sends $W$ back into itself. We finish up by picking some eigenvector $v_0\in W$ of $z$, which we know must exist because we’re working over an algebraically closed field. Incidentally, we can then extend $\lambda$ to all of $L$ by using $z(v_0)=\lambda(z)v_0$.

August 25, 2012 Posted by | Algebra, Lie Algebras | 1 Comment