The Unapologetic Mathematician

Mathematics for the interested outsider

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 | Algebra, Differential Topology, Lie Groups, Topology

2 Comments »

  1. [...] 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 [...]

    Pingback by Seventh Linkfest | June 13, 2011 | Reply

  2. [...] 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 . [...]

    Pingback by The Adjoint Representation « The Unapologetic Mathematician | June 13, 2011 | Reply


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