Calculating the Lie algebra of a Lie group

Posted on 6 April, 2015 by jesperg

The Lie algebra of a Lie group is the tangent space at the identity. Thus we can compute the Lie algebra by differentiating the Lie group and evaluating the derivative at the identity.

Example 1. Let G=SL(2). These are the 2\times 2 matrices with unit determinant:

\displaystyle\begin{pmatrix}x&y\\z&w\end{pmatrix},\qquad xw-yz=1

Differentiating, we obtain:

\displaystyle\begin{pmatrix}dx&dy\\dz&dw\end{pmatrix},\qquad wdx+xdw-zdy-ydz=0

and setting x=w=1, y=z=0 gives

\displaystyle\begin{pmatrix}dx&dy\\dz&dw\end{pmatrix},\qquad dx+dw=0,

which is simplified into

\displaystyle\begin{pmatrix}dx&dy\\dz&-dx\end{pmatrix}.

Thus the we have shown that the Lie algebra of SL(2) is the set of  2\times2 traceless matrices.

Example 2. Let G=SO(n). We have the relationship AA^T=I, which when differentiated gives dA\,A^T+A\,dA^T=0. Evaluated at A=I, this is dA=-dA^T and thus the Lie algebra consists of the antisymmetric matrices.

Example 3. Let G=PGL(n). The group consists of equivalence classes of matrices differing by multiplication by a scalar: A\sim cA, c\in\mathbf{R}\setminus 0. Differentiating gives  dA\sim c\,dA+dc\,A, dc\in\mathbf{R}, which when evaluated at A=I, c=1 gives that the Lie algebra is the equivalence classes dA\sim dA+dcI, dc\in\mathbf{R}.

Example 4. Let G=SL(n). We have that \det A=1. By differentiating and using Jacobi’s formula d\det(A)=\mathrm{tr}(\mathrm{adj}(A)dA)  we get that \mathrm{tr}(\mathrm{adj}(A)dA) = 0, which when evaluated at A=I gives \mathrm{tr}(dA)=0 .

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