5
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So, I know the following facts:

  1. A Lie group is a group and a Manifold, whose group structure is continuous with respect to the Manifold structure
  2. The Lie algebra is a vector space with a Lie bracket structure on it.

  3. Every Lie group has a corresponding Lie algebra.

  4. The Lie algebra represents the "infinitesimal behaviour" of the Lie group.

Till here, stuff makes sense. However, this is where I lose it:

  1. The Vector space of all Vector fields over a Manifold form a Lie algebra with the Lie bracket of vector fields structure.

Why do I care about fact 5? Is it because it is "cute" that the Lie bracket exists? What do I gain by showing that vector fields have a lie bracket structure?

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    $\begingroup$ If the manifold is a Lie group, then you can take the left-invariant vector fields, show that these are closed under the bracket structure on all vector fields on the manifold, and this will be the Lie algebra of the group. I guess that's cute? $\endgroup$ – Joppy Apr 30 '18 at 14:39
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    $\begingroup$ en.m.wikipedia.org/wiki/… $\endgroup$ – Lorenzo Najt Apr 30 '18 at 15:54
  • $\begingroup$ @Joppy "...this will be the Lie algebra of the group" - Is that not the definition of the Lie algebra of the Lie group? I was asking more about the Lie algebra of a general manifold (which is not a Lie group) $\endgroup$ – Siddharth Bhat Apr 30 '18 at 17:58
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    $\begingroup$ It's a definition of the Lie algebra of a Lie group, but another one is for example the tangent space at the identity, with the bracket induced by the differential of the conjugation map. $\endgroup$ – Joppy May 1 '18 at 0:52
  • $\begingroup$ @Joppy - could you please expand on this comment with an answer? I have not seen this equivalent definition, and would be nice if you could expand on what you said :) $\endgroup$ – Siddharth Bhat May 2 '18 at 9:16
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I can't answer the question in your title -- maybe you just don't care. ;-)

But I'll tell you some reasons why I find it interesting and useful.

  1. First, and perhaps deepest, is that you can view the group of all diffeomorphisms of a smooth manifold $M$ as an infinite-dimensional Fréchet Lie group, and its Lie algebra is exactly $\mathfrak X(M)$ with its Lie bracket structure (see this article by Richard Hamilton for a beautiful exposition of this point of view).

  2. Second, any smooth right action by a (finite-dimensional) Lie group $G$ on $M$ determines a Lie algebra homomorphism from $\operatorname{Lie}(G)$ to $\mathfrak X(M)$, sending $X\in \operatorname{Lie}(G)$ to the vector field $\widehat X\in \mathfrak X(M)$ defined by $$ \widehat X_p = \left.\frac{d}{dt}\right|_{t=0} p\cdot \exp tX. $$ Conversely, given any finite-dimensional Lie subalgebra $\mathfrak g\subset \mathfrak X(M)$ with the property that every vector field in $\mathfrak g$ is complete, there is a smooth right action of the simply connected Lie group $G$ whose Lie algebra is isomorphic to $\mathfrak g$, and $\mathfrak g$ is the image of the Lie algebra homomorphism described above. (See pp. 525-530 of my Introduction to Smooth Manifolds [ISM].)

  3. You can think of the Lie algebra structure of $\mathfrak X(M)$ as providing obstructions to commutativity of flows: Two smooth flows on $M$ commute if and only if their infinitesimal generators have zero Lie bracket [ISM, Thm. 9.44].

  4. If $M$ is endowed with a complete Riemannian metric, the set of all Killing vector fields is a finite-dimensional Lie subalgebra of $\mathfrak X(M)$, which is the Lie algebra of the full isometry group of $M$.

  5. In the presence of a symplectic structure, there is a map from $C^\infty(M)$ to $\mathfrak X(M)$ sending $f$ to its Hamiltonian vector field $X_f$, which gives a Lie algebra isomorphism (or anti-isomorphism, depending on your conventions) between $C^\infty(M)/\{\text{constants}\}$ with its Poisson bracket structure and a certain Lie subalgebra $\mathscr H(M)\subseteq \mathfrak X(M)$, the algebra of Hamiltonian vector fields. This is in turn a subalgebra of a larger Lie subalgebra $\mathscr S(M)\subseteq \mathfrak X(M)$, the symplectic vector fields, and the quotient $\mathscr S(M)/\mathscr H(M)$ is naturally isomorphic to the first de Rham cohomology of $M$. (See, for example, [ISM, Chap. 22].) The Lie algebra structure of $\mathscr H(M)$ plays a central role in dynamical systems, for example in identifying completely integrable systems and in Noether's theorem about the relationship between symmetries and conserved quantities.

  6. The Frobenius theorem says that a smooth distribution (i.e., vector subbundle) $D\subseteq TM$ is tangent to a foliation if and only if the set of smooth sections of $D$ is a Lie subalgebra of $\mathfrak X(M)$. [Thanks to Jason deVito for suggesting this one.]
  7. And of course I shouldn't leave out the one most directly related to Lie groups -- if $G$ is a Lie group, then the Lie algebra of $G$ is naturally realized as the Lie subalgebra $\mathfrak g\subset \mathfrak X(G)$ consisting of left-invariant vector fields. [Pointed out by @Joppy in comments above.]

I'm convinced -- are you?

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  • $\begingroup$ Wow, thanks for this in-depth answer! In particular, (3) is a direct, intuitive reason for caring. (1) is a deep, crazy reason :) $\endgroup$ – Siddharth Bhat May 2 '18 at 21:00
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    $\begingroup$ Oh wow, I'm reading your book. Thanks a lot for writing it, it's very well written (I love the exposition), and the exercises are fun to do as well! $\endgroup$ – Siddharth Bhat May 2 '18 at 21:01
  • $\begingroup$ @SiddharthBhat: You're welcome! And thanks for the kind compliment. $\endgroup$ – Jack Lee May 2 '18 at 21:03
  • $\begingroup$ I would only add one thing to this fantastic answer: The Frobenius theorem tells you exactly when a distribution comes from a submanifold, and it's tied directly to the Lie algebra strcture on $\mathfrak X(M)$. $\endgroup$ – Jason DeVito May 2 '18 at 21:15
  • $\begingroup$ @JasonDeVito: Great suggestion! I'll add that to my list. $\endgroup$ – Jack Lee May 2 '18 at 21:27

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