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In my lecture, we gave a very sloppy (physics people ...) proof of the fact that the Lie algebra $\mathfrak{g}$ of a matrix Lie group $G$ is a subspace of $\text{Mat}_n(\mathbb{F})$. I am not satisfied and I would love to understand this fact in the language of differential geometry that I am learning right now.

What I know:

The Lie algebra $\mathfrak{g}$ of a matrix Lie group $G$ is the set of all left-invariant vector fields on $G$. This set is isomorphic to the tangent space at the identity of the Lie group $G$. The elements $v$ of the tangent space at the identity are according to my definition of the tangent space real-valued functions $$v:F(G)\to \mathbb{R}$$ where $F(M)$ is the set of all smooth-real valued functions on the manifold $G$. The tangent vectors are $\mathbb{R}$ linear and fulfil a Leibniz rule and the tangent space at $p$, namely $T_pG$ is simply the set of all these tangent vectors.

What I want to know

Given the above definitions, how can I formally understand that the members of the tangent space at the identity of a matrix Lie group are matrices?

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A matrix Lie group is a closed subgroup of $\text{GL}(n,\mathbb{K})$ where $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$

Now $\text{GL}(n,\mathbb{K})$ is an open subset of the vector space $\mathbb{K}^{n\times n},$ and you may use:

  1. If $V$ is a vector space of dimension $n,$ any norm in $V$ determines a topology independent of the choice of the norm, and $V$ has a standard smooth structure making it into an $n$-dimensional manifold.
  2. If $V$ is a vector space with its standard manifold structure, the tangent space at a point $p\in V$ is naturally isomorphic to $V,$ that is, there is a canonical isomorphism $\varphi_V:V \to T_pV:$ This isomorphism is such that if $L:V\to W$ is a linear map between two vector spaces then $dL_p\circ \varphi_V=\varphi_W\circ L:V\to T_{Lp}W$ where $dL_p:T_pV\to T_{Lp}W$ is the differential of $L$ at $p.$ This allow us to identify the tangent space $T_pV$ with $V$ itself.
  3. If $\iota:N\to M$ is the inclusion of a submanifold $N\subset M$ and $p\in N$ then $d\iota_p:T_pN\to T_pM$ gives a inclusion of $T_pN$ into $T_pM.$ If $N$ is an open submanifold of $M$ then $d\iota_p$ is an isomorphism, so we can identify $T_pN$ with $T_pM.$

Now let $G\subset \text{GL}(n,\mathbb{K})$ be a matrix Lie group and $I\in G$ be the identity. By 2, the tangent space $T_I(\mathbb{K}^{n\times n})$ is identified with $\mathbb{K}^{n\times n}$ itself. By 3, the tangent space $T_I(\text{GL}(n,\mathbb{K}))$ can be identified with $T_I(\mathbb{K}^{n\times n}),$ hence with $\mathbb{K}^{n\times n}.$ Hence by the first statement in 3, the tangent space $T_IG$ is a subspace of $\mathbb{K}^{n\times n}.$

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  • $\begingroup$ Hey, I really like your answer. It has exactly the stuff in it, I just learned in differential geometry. To clarify, I understand that by 3, you can identify $T_I(GL(n))$ with $T_I(\mathbb{R}^{nxn})$ but then to argue about the closed subgroup $G$ you cannot use 3 again, can you? Or am I misunderstanding? $\endgroup$ – Marsl Dec 13 '18 at 10:48
  • $\begingroup$ I'm using 3 again but only the first sentence: Since $G$ is a submanifold of $\text{GL}(n,\mathbb{R})$ then $T_I(G)$ is a subspace of $T_I(\text{GL}(n,\mathbb{R})).$ $\endgroup$ – positrón0802 Dec 13 '18 at 12:09
  • $\begingroup$ I see, nice. One more question. What about matrix groups like $U(n)$ which is actually a closed subgroup of $GL(n,\mathbb{C})$. Now, I feel, I cannot argue the same way, since the $\mathbb{C}^{n \times n}$ is a complex finite-dimensional vector space and thus not a manifold, so the whole argument via canonical isomorphisms doesnt work? $\endgroup$ – Marsl Dec 13 '18 at 12:14
  • $\begingroup$ $\mathbb{C}^{n\times n}$ is also a real vector space, of dimension $(2n)^2,$ and is in fact diffeomorphic to $\mathbb{R}^{2n\times 2n}$ as a (real) manifold, since $\mathbb{C}\cong \mathbb{R}^{2n}$ as topological spaces (with the standard topology). $\endgroup$ – positrón0802 Dec 13 '18 at 12:21
  • $\begingroup$ You just make me realize that I should replace $\mathbb{R}$ by $\mathbb{K}$ where $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$ in most parts of my answer, I'll edit it. $\endgroup$ – positrón0802 Dec 13 '18 at 12:26
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You have to use another characterization of the tangent space (which is in natural bijection with your definition, alhough it is not wholly straightforward), that is:

the tangent space of a manifold $M$ at a point $x$ is the quotient $S/\sim$, where $S$ is the set of all smooth paths $\gamma:[-a,b] \rightarrow M$ mapping $0$ to $x$, where $a,b>0$, and $\gamma_1 \sim \gamma_2$ iff in some chart around $x$ (ie in every chart around $x$) $\gamma_1’(0)=\gamma_2’(0)$.

The bijection is given by $\gamma \longmapsto (f \longmapsto d/dt(f\circ\gamma) (t=0)$.

With this alternate definition, when your manifold is embedded in some vector space, the tangent space at some point is canonically embedded in the same vector space.

Edit:

0) I am using “derivations” as the name of tangent vectors according to your definition. 1) The bijection is defined from the space of equivalence classes of paths to derivations of $F(M)$ at $x$: to the path $\gamma$ you associate the derivation $D_{\gamma}: f \longmapsto \frac{d}{dt}_{|t=0}(f \circ \gamma)$.

You can for instance take $M=\mathbb{R}^n$: then the equivalence classes are exactly given by the (usual linear) derivative of paths at $t=0$, and to the path $\gamma$ you associate the derivation $D_{\gamma}: f \longmapsto d_x f(\gamma’(0))$.

2) When your manifold is embedded in a vector space $E$ (with a finite dimension), then two paths are equivalent iff they have same linear derivative at $t=0$ (as smooth functions from an interval to $E$). So you can represent each equivalence class in the tangent space by the underlying common derivative of the paths at $t=0$, which is a vector from $E$, and you get an embedding $T_x M \rightarrow E$. Now, when $M$ is a matrix group, it is embedded in the space of square matrices, thus so are ots tangent spaces.

3) The definition of the tangent space using derivations is pretty clean and satisfying (not that your physics people would care ;) ). It can however sometimes be inconvenient (when proving, say, that the tangent bundle exists as a manifold).

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  • $\begingroup$ Thanks for the answer. I kind of hoped for an answer that would circumvent the definition in terms of tangents to curves as I feel I dont understand it very well. Could you expand on how the $\gamma'$ can be seen as matrices, as I dont really understand your explanation/the bijection you mention. $\endgroup$ – Marsl Dec 13 '18 at 0:45
  • $\begingroup$ Hey, thanks for the edit. If I understand your answer correctly now, the answer to my question lies really in fact that my manifold is embedded in a vector space E whose tangent spaces is isomorphic to E and thus I also have an embedding $T_x M\to E$ as you write. I might miss something, but where is the explicit definition of the tangent spaces (be it in terms of derivations or tangents to curves) even important? $\endgroup$ – Marsl Dec 13 '18 at 11:10
  • $\begingroup$ Not really, but at some rather close point you will have to give the set of all tangent spaces a manifold structure (you get the tangent bundle). And it is somewhat easier if you consider tangent paths. $\endgroup$ – Mindlack Dec 13 '18 at 11:35
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Let $\gamma$ be a curve in the matrices; that is $\gamma:(-\epsilon,\epsilon) \rightarrow G$ such that $\gamma(0)$ = $I$. We want to know the possible tangent vectors to these curves. From what everyone has said, we can view this as a curve into $\mathbb{R}^{n^2}$. We know from calculus, that every v $\in \mathbb{R}^{n^2}$ is a possible tangent vector. So it may be easier to see that every matrix is a tangent vector when we view $\gamma$ as a matrix curve.

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