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.

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June 9, 2011 - Posted by John Armstrong | Algebra, Differential Topology, Lie Groups, Topology

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[...] John Armstrong: Integral curves and local flows, The Maximal Flow of a Vector Field, Vector fields on compact manifolds are complete, Maps Intertwining Vector Fields, The Lie algebra of a Lie group, General linear groups are Lie groups, The Lie algebra of a general linear group [...]

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[...] get even more specific, we can consider the adjoint representation of on its Lie algebra . I say that is just itself. That is, if we view as an open subset of then we can identify . [...]

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