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EDIT(3): I am looking for a basis for the Lie algebra of polynomial vector fields on $S^3$.

EDIT(2): I am fairly certain now that my question is more along the lines of, what does the Lie algebra of polynomial vector fields on $S^3$ look like?

EDIT: my question is really the following. The Lie algebra of the diffeomorphism group of $S^1$ is the Witt algebra. What is the corresponding Lie algebra for $S^3$?

(1) What is the diffeomorphism group of the 3-sphere? My reason for asking is that I want to know if there is an analogue of the Witt (=centerless Virasoro) algebra in three dimensions. I am aware of the $W_n$ series in Cartan's classification, but this is not the generalization I am looking for.

(1.0) Alternatively/equivalently, I would like to know how to describe "regular" (smooth?) sections of the tangent bundle on $S^3$ - it seems like it might be helpful (to me, at least) to think of the fibers as copies of $sl(2)(\cong su(2))$. It's been a while since I looked at principal fiber bundles, but this definitely reminds me of one.

(0) Currently I'm trying to think of all of this stuff in terms of [unit] quaternions. This seems promising to me for a number of reasons, so if you can tell me anything about the above in such terms or point me towards papers/preprints/steles about the above in quaternionic language, that would be fantabulous.

(Obligatory "sorry if this is vague to the point of madness" and "sorry if this has been answered previously; I did my best to check the related questions.")

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    $\begingroup$ The tangent bundle of $S^3$ is trivial (view $S^3$ as the group of unit quaternions, and use the group law to trivialize the bundle) so the smooth sections are $C^\infty$ maps $S^3\rightarrow\mathbb{R}^3$. $\endgroup$
    – Alain Valette
    Commented Aug 7, 2011 at 19:58
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    $\begingroup$ I suspected [something like] this. That's why I'm interested in $C^{\infty}$ maps $S^3\rightarrow S^3(\cong\mathbb{R}^3\cup\{\infty\})$. Thanks for the bundle clarification. Very clear. $\endgroup$
    – Daniel Fleisher
    Commented Aug 7, 2011 at 20:08
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    $\begingroup$ In fact, $S^3$ is a Lie group. One can find three linearly-independent and nonvanishing vector fields, trivializing $TS^3$ and forming a basis for the corresponding Lie algebra, see en.wikipedia.org/wiki/3-sphere $\endgroup$
    – Francesco Polizzi
    Commented Aug 7, 2011 at 20:10
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    $\begingroup$ I think that I now understand what your question really is. You have the Lie algebra of polynomial vector fields on $S^3$, which I defined in my answer. You'd like to know if there is a way of describing it in a "concrete" way (i.e., in the way the Witt algebra is usually defined: by formulas). So your question could be formulated as: "could someone provide a nice basis of the Lie algebra of polynomial vector fields on $S^3$ in which the Lie bracket can be described by a nice closed formula?". $\endgroup$
    – André Henriques
    Commented Aug 7, 2011 at 21:21
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    $\begingroup$ Yes, that's pretty much what I want. The fact that I haven't found anything resembling this makes me somewhat pessimistic. $\endgroup$
    – Daniel Fleisher
    Commented Aug 7, 2011 at 22:08

3 Answers 3

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You mention the Witt algebra, which is a dense Lie algebra inside the Lie algebra of smooth vector fields on $S^1$...
There is a similar dense Lie algebra inside the Lie algebra of smooth vector fields on $S^3$: the Lie algebra of polynomial vector fields. Since it might not be clear what one means by "polynomial vector fields", let me be precise: these are algebraic vector fields on the complexification $S^3_\mathbb C=SU(2)_\mathbb C=SL(2,\mathbb C)$, subject to the reality condition that says that their restriction to $S^3$ is everywhere tangent to $S^3$.

But you also mention the diffeomorphism group of $S^3$. I am not aware of any dense subgroup of $Diff(S^3):=Diff_{smooth}(S^3)$ that would be to $Vect_{polynomial}(S^3)$ the same as $Diff_{smooth}(S^3)$ is to $Vect_{smooth}(S^3)$.

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  • $\begingroup$ I feel rather stupid - I never meant to say anything about diffeomorphisms. I am interested in the Lie algebra of polynomial vector fields on $S^3$. Editing the post again... (sorry) How does one work with the algebraic vector fields you mentioned - that is, how does one impose the reality condition to get a nice basis for the Lie algebra? $\endgroup$
    – Daniel Fleisher
    Commented Aug 7, 2011 at 20:51
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    $\begingroup$ You may forget the reality condition that I mentioned in my answer: the Witt algebra is the complexified Lie agebra of vector fields on $S^1$. You seem to care about the complexified Lie algebra of vector fields on $S^3$. $\endgroup$
    – André Henriques
    Commented Aug 7, 2011 at 21:23
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Complementing André's answer, here's another possible definition of polynomial vector fields on $S^3$. You think of $S^3$ as the unit sphere in $\mathbb{R}^4$ and consider polynomial vector fields on $\mathbb{R}^4$ which are tangent to the sphere; that is, which annihilate the function $\sum x_i^2$.

One thing to point out, which may or may not be relevant to the applications you have in mind but which I mention since you did mention the Virasoro algebra, is that the structure of the diffeomorphism algebras (or algebras of polynomial vector fields) in dimension greater than 1 is very different than in dimension 1. For example, you don't have a nice decomposition such as the one

$$ \mathfrak{Vir} = \mathfrak{Vir}^- \oplus \mathfrak{Vir}^0 \oplus \mathfrak{Vir}^+ $$ for the Virasoro algebra, and this in turn hinders the construction of positive energy representations.

I am aware on some work on this topic in the mathematical physics literature; e.g.,

Fock space representations of the algebra of diffeomorphisms of the $N$-torus, F Figueirido and E Ramos

and also papers by TA Larsson.

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    $\begingroup$ @ José: The fact that our two notions of "polynomial vector fields on $S^3$" agree is actually non-trivial. But this is indeed the case. The reason is the following: the variety $S^3_{\mathbb C}=\{(x,y,z,u)\in\mathbb C^4|x^2+y^2+z^2+u^2=1\}$ is isomorphic to the Lie group $SL(2,\mathbb C)=\{(a,b,c,d)\in\mathbb C^4|ad-bc=1\}$. $\endgroup$
    – André Henriques
    Commented Aug 8, 2011 at 19:39
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The "Smale Conjecture" (a theorem of Hatcher http://www.jstor.org/pss/2007035) says that the natural inclusion $\operatorname{O}(4)\hookrightarrow\operatorname{Diff}(\mathbb S^3)$ is a homotopy equivalence. Perhaps this is along the lines of what you are looking for.

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    $\begingroup$ I think I've asked my question[s] terribly. I'm going to edit the initial post. $\endgroup$
    – Daniel Fleisher
    Commented Aug 7, 2011 at 20:33

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