# The Unapologetic Mathematician

## The Lie Algebra of a General Linear Group

Since $GL_n(\mathbb{R})$ is an open submanifold of $M_n(\mathbb{R})$, the tangent space of $GL_n(\mathbb{R})$ at any matrix $A$ is the same as the tangent space to $M_n(\mathbb{R})$ at $A$. And since $M_n(\mathbb{R})$ is (isomorphic to) a Euclidean space, we can identify $M_n(\mathbb{R})$ with $\mathcal{T}_AM_n(\mathbb{R})$ using the canonical isomorphism $\mathcal{I}_A:M_n(\mathbb{R})\to\mathcal{T}_AM_n(\mathbb{R})$. In particular, we can identify it with the tangent space at the identity matrix $I$, and thus with the Lie algebra $\mathfrak{gl}_n(\mathbb{R})$ of $GL_n(\mathbb{R})$:

$\displaystyle M_n(\mathbb{R})\cong\mathcal{T}_IGL_n(\mathbb{R})\cong\mathfrak{gl}_n(\mathbb{R})$

But this only covers the vector space structures. Since $M_n(\mathbb{R})$ is an associative algebra it automatically has a bracket: the commutator. Is this the same as the bracket on $\mathfrak{gl}_n(\mathbb{R})$ under this vector space isomorphism? Indeed it is.

To see this, let $A$ be a matrix in $M_n(\mathbb{R})$ and assign $X(I)=\mathcal{I}_I(A)\in\mathcal{T}_IGL_n(\mathbb{R})$. This specifies the value of the vector field $X$ at the identity in $GL_n(\mathbb{R})$. We extend this to a left-invariant vector field by setting

$\displaystyle X(g)=L_{g*}X(I)=L_{g*}\mathcal{I}_I(A)=\mathcal{I}_{g}(gA)$

where we subtly slip from left-translation by $g$ within $GL_n(\mathbb{R})$ to left-translation within the larger manifold $M_n(\mathbb{R})$. We do the same thing to go from another matrix $B$ to another left-invariant vector field $Y$.

Now that we have our hands on two left-invariant vector fields $X$ and $Y$ coming from two matrices $A$ and $B$. We will calculate the Lie bracket $[X,Y]$ — we know that it must be left-invariant — and verify that its value at $I$ indeed corresponds to the commutator $AB-BA$.

Let $u^{ij}:GL_n(\mathbb{R})\to\mathbb{R}$ be the function sending an $n\times n$ matrix to its $(i,j)$ entry. We hit it with one of our vector fields:

$\displaystyle Yu^{ij}(g)=Y_gu^{ij}=\mathcal{I}_g(gB)u^{ij}=u^{ij}(gB)$

That is, $Yu^{ij}=u^{ij}\circ R_B$, where $R_B$ is right-translation by $B$. To apply the vector $X_I=A$ to this function, we must take its derivative at $I$ in the direction of $A$. If we consider the curve through $I$ defined by $c(t)=I+tA$ we find that

$\displaystyle X_IYu^{ij}=\dot{c}(0)(u^{ij}\circ R_B)=\frac{d}{dt}(u^{ij}(B+tAB))\Big\vert_{t=0}=(AB)_{i,j}$

Similarly, we find that $Y_IXu^{ij}=(BA)_{i,j}$. And thus

$\displaystyle [X,Y]_Iu^{ij}=(AB-BA)_{i,j}=\mathcal{I}_I(AB-BA)(u^{ij})$

Of course, for any $Q\in M_n(\mathbb{R})$ we have the decomposition

$\displaystyle\mathcal{I}_IQ=\sum\limits_{i,j=1}^n\mathcal{I}_IQ(u^{ij})\frac{\partial}{\partial u^{ij}}\Big\vert_I$

Therefore, since we’ve calculated $[\mathcal{I}_IA,\mathcal{I}_IB](u^{ij})=\mathcal{I}_I(AB-BA)(u^{ij})$ we know these two vectors have all the same components, and thus are the same vector. And so we conclude that the Lie bracket on $\mathfrak{gl}_n(\mathbb{R})$ agrees with the commutator on $M_n(\mathbb{R})$, and thus that these two are isomorphic as Lie algebras.

June 9, 2011

## General Linear Groups are Lie Groups

One of the most important examples of a Lie group we’ve already seen: the general linear group $GL(V)$ of a finite dimensional vector space $V$. Of course for the vector space $\mathbb{R}^n$ this is the same as — or at least isomorphic to — the group $GL_n(\mathbb{R})$ of all invertible $n\times n$ real matrices, so that’s a Lie group we can really get our hands on. And if $V$ has dimension $n$, then $V\cong\mathbb{R}^n$, and thus $GL(V)\cong GL_n(\mathbb{R})$.

So, how do we know that it’s a Lie group? Well, obviously it’s a group, but what about the topology? The matrix group $GL_n(\mathbb{R})$ sits inside the algebra $M_n(\mathbb{R})$ of all $n\times n$ matrices, which is an $n^2$-dimensional vector space. Even better, it’s an open subset, which we can see by considering the (continuous) map $\mathrm{det}:M_n(\mathbb{R})\to\mathbb{R}$. Since $GL_n(\mathbb{R})$ is the preimage of $\mathbb{R}\setminus\{0\}$ — which is an open subset of $\mathbb{R}$$GL_n(\mathbb{R})$ is an open subset of $M_n(\mathbb{R})$.

So we can conclude that $GL_n(\mathbb{R})$ is an open submanifold of $M_n$, which comes equipped with the standard differentiable structure on $\mathbb{R}^{n^2}$. Matrix multiplication is clearly smooth, since we can write each component of a product matrix $AB$ as a (quadratic) polynomial in the entries of $A$ and $B$. As for inversion, Cramer’s rule expresses the entries of the inverse matrix $A^{-1}$ as the quotient of a (degree $n-1$) polynomial in the entries of $A$ and the determinant of $A$. So long as $A$ is invertible these are two nonzero smooth functions, and thus their quotient is smooth at $A$.

June 9, 2011