I'm looking for an intuitive explanation of the meaning of the Lie algebra for a matrix Lie group from a differential geometry perspective.

Right now, the procedure I've been following is using the Regular Level Set Theorem and showing that for some surjective function $F : M \rightarrow N$, if $c$ is a regular value of $F$ then $F^{-1}(c) := \{ p \in M : F(p) = c\}$ is an embedded submanifold and $ Lie(F^{-1}(c)) = T_𝟙 F^{-1}(c) \cong Ker(F_{*, 𝟙})$.

I've been able to show that the lie algebra of the orthogonal group:

$$ O(n) = \{ A \in M_{n \times n}(\mathbb{R}) : AA^T = 𝟙 \} $$ is $Skew(n, \mathbb{R}) = \{ A \in M_{n \times n}(\mathbb{R}) : A^T = - A\}$, and that the Lie algebra of the Special Linear group:

$$ SL(n, \mathbb{R}) = \{ A \in M_{n\times n} : det(A) = 1 \} $$

is the set of trace $0$ matricies.

The Lie algebras I'm getting seem completely arbitrary to me and I can't really see what the relation is with the original Lie group!

  • $\begingroup$ The Lie algebra of a Lie group is just the tangent space at the identity. $\endgroup$ – Lord Shark the Unknown Dec 10 '18 at 4:58
  • $\begingroup$ @LordSharktheUnknown Yes, but specifically I want to know how the Lie algebra of the above examples relates to the matrix group they come from. I found them by using the fact that the tangent space at the identity is isomorphic to the Kernel of the pushforward of $F$ at the identity, but just using that definition isn't very enlightening $\endgroup$ – David Feng Dec 10 '18 at 5:04
  • $\begingroup$ The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation. $\endgroup$ – Randall Dec 10 '18 at 5:21

Maybe you don't know there's an exponential map that takes the Lie algebra to the Lie group: $$\exp \colon \mathfrak{g} \to G$$

In general, this gives a diffeomorphism from a neighborhood of 0 in the Lie algebra to a neighborhood of the identity in $G$.

For your concrete matrix groups, we can use this to compute the Lie algebras easily. Let's look at the case of $O(n)$. Say $X$ is in the Lie algebra. We have the 1-parameter subgroup $e^{tX}$ which satisfies $e^{tX}e^{tX^T} = I$. Differentiating at $t = 0$, we get $X+X^T = 0$ as you discovered. You can do the $SL(n)$ case just as easily.

  • 3
    $\begingroup$ Another example that I found illuminating is the computation of $$\tag{1}\exp t\begin{bmatrix} 0 & -1 \\ 1&0\end{bmatrix},$$ using the definition $$\exp(A)=I+A+\frac12 A^2 +\ldots $$ The Lie algebra of $SO(2)$ is the linear span of $\begin{bmatrix} 0 & -1 \\ 1&0\end{bmatrix}$, and indeed (1) magically yields $$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix}.$$ $\endgroup$ – Giuseppe Negro Dec 10 '18 at 8:57

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