# Is there a unique "natural" action of $\mathsf{SL}_{n+1}$ on $\mathbb{R}^n$?

## Context

By acting naturally via $$\mathsf{SL}_3$$ on $$\mathbb{RP}^2=\{[x:y:z]\}$$ and by taking the induced action on the affine hyperplane $$z=1$$ (which we identify with $$\mathbb{R}^2$$), one can realize the corresponding Lie algebra $$\mathfrak{sl}_3$$ as a sub-algebra of the algebra $$\mathfrak{X}(\mathbb{R}^2)$$ of vector fields of $$\mathbb{R}^2$$: by carrying out this simple exercise, one ends up with 8 vector fields, 6 of which are linear, and two contain quadratic coefficients, namely $$X_A=-xy\partial_x-y^2\partial_y\, ,\quad X_B=-x^2\partial_x-xy\partial_y\, ,$$ that correspond to the matrices $$A=\left(\begin{array}{ccc}0&0&0\\ 0&0&0\\0&1&0\end{array}\right)\,,\quad B=\left(\begin{array}{ccc}0&0&0\\ 0&0&0\\1&0&0\end{array}\right)\, ,$$ respectively. Image below shows how to use the flow of $$X\in\mathfrak{sl}_3$$ to obtain a vector field on the plane $$z=1$$:

The 6 linear vector fields are of course complete, whereas the flows of $$X_A$$ and $$X_B$$ display a singularity at $$t=-\tfrac{1}{y}$$ and $$t=-\tfrac{1}{x}$$, respectively. This remark motivated the first question:

For dimensions reasons (as observed in the comments), for every $$n\ge 1$$ there is no embedding $$\mathfrak{sl}_{n+1}\hookrightarrow\mathfrak{X}(\mathbb{R}^n)\, ,$$ whose image is made of linear (allowing a constant) vector fields.

However for any $$n\geq 1$$ it is easy to find an embedding $$i_{\textrm{nat}}:\mathfrak{sl}_{n+1}\hookrightarrow\mathfrak{X}(\mathbb{R}^n)\, ,$$ which we may call natural, because it is induced by the natural action of $$\mathsf{SL}_{n+1}$$ on $$\mathbb{RP}^n$$ by the same construction as above, one may wonder the following (initially "Question 2"):

## Question

Given an arbitrary embedding $$i :\mathfrak{sl}_{n+1}\hookrightarrow\mathfrak{X}(\mathbb{R}^n)\, ,$$ it is true that $$i$$ must necessarily be equivalent to $$i_{\textrm{nat}}$$? If not, which conditions must $$i$$ satisfy, to be such?

In other words, one has a collection $$X_1,X_2,\ldots, X_{d(n)}$$, where $$d(n)=(n+1)^2-1=\dim \mathsf{SL}_{n+1}$$ of vector fields on $$\mathbb{R}^n$$, that commute according to the commutation relations of $$\mathsf{SL}_{n+1}$$ and the question can be recast as follows: is there a diffeomorphism of $$\mathbb{R}^n$$ that transforms the aforementioned vector fields into the images via $$i_{\textrm{nat}}$$ of a suitable set of generators of $$\mathfrak{sl}_{n+1}$$? What kind of obstruction (if any) one should expect?

• I am confused about your question 1. How do you embed a vector space of dimension $8$ into a vector space of dimension $4$? Jun 10, 2021 at 9:54
• For question 2, a good start would probably be to ask if $gl_n\subset sl_{n+1}$ can be embedded in vector fields on $\mathbb{R}^n$ in an essentially unique way. Jun 10, 2021 at 9:55
• About your confusion about question 1, my bad: by linear I meant linear or constant (I.e., of degree less or equal to one). In any case, there is only 2+4=6 of them, so that pure dimensional considerations are enough. The true question is the second one. Jun 10, 2021 at 10:13
• Perhaps one can look at the "isotropy" of a point. In this case, it is a parabolic sub-algebra associated with a point of $\mathbb{P}^n$. It is possible that there are other sub-algebras of index $n$ in the lie algebra $\mathfrak{sl}_{n+1}$. Jun 10, 2021 at 11:11
• I think that VladimirDotsenko in his second remark had in mind the parabolic sub-algebra $\mathfrak{p}=\mathfrak{gl}_n\oplus\mathbb{R}^n$ of index $n$ mentioned in the comment of @Kapil . If I got him right, if there is a point $x\in \mathbb{R}^n$, such that the kernel of $i_x:\mathfrak{sl}_{n+1}\to T_x\mathbb{R}^n$ is isomorphic to $\mathfrak{p}$, then there is hope of "rectifying" the embedding $i$ into $i_{\textrm{nat}}$. This looks like a good necessary condition; actually constructing the "rectifying diffeomorfism" is another story. Jun 11, 2021 at 4:31

There are two aspects to this question, the global question and the local question. Also, the case $$n=1$$ is different from $$n>1$$. Basically, the answer is 'essentially yes, but with some caveats'.

Here's a sample of the kind of results one might consider as an answer in the case $$n=1$$:

First, an old theorem of Sophus Lie:

Theorem: (Lie) Let $$L\subset\mathfrak{X}(\mathbb{R}^1)$$ be a Lie algebra of vector fields that is locally transitive, i.e., for every $$p\in\mathbb{R}$$, there is a vector field $$X\in L$$ such that $$X(p)\not=0$$. Then either

1. $$\dim L=1$$ and, up to local diffeomorphism, $$L$$ is spanned by $$\frac{\partial}{\partial x}$$,

2. $$\dim L=2$$ and, up to local diffeomorphism, $$L$$ is spanned by $$\frac{\partial}{\partial x},\ x\,\frac{\partial}{\partial x}$$,

3. $$\dim L=3$$ and, up to local diffeomorphism, $$L$$ is spanned by $$\frac{\partial}{\partial x},\ x\,\frac{\partial}{\partial x},\ x^2\,\frac{\partial}{\partial x}$$, or

4. $$\dim L=\infty$$, and for every $$k$$ and every $$p\in\mathbb{R}$$, there is a vector field $$X_k\in L$$ such that $$X_k$$ vanishes to order exactly $$k$$ at $$p$$.

Then there is a global result that is not too hard to prove:

Theorem: Suppose that $$L\subset\mathfrak{X}(\mathbb{R})$$ is a locally transitive Lie algebra of dimension $$3$$. Then there is a diffeomorphism $$u:\mathbb{R}\to u(\mathbb{R})\subseteq\mathbb{R}$$ that carries $$L$$ into the span of the restrictions to $$u(\mathbb{R})$$ of the vector fields $$\frac{\partial}{\partial x}$$, $$\cos x\,\frac{\partial}{\partial x}$$, and $$\sin x\,\frac{\partial}{\partial x}$$.

Note that all the vector fields in $$L$$ are complete if and only if $$u(\mathbb{R})=\mathbb{R}$$. In particular, if $$L\subset\mathfrak{X}(\mathbb{R})$$ is the span of $$\frac{\partial}{\partial x}$$, $$\cos x\,\frac{\partial}{\partial x}$$, and $$\sin x\,\frac{\partial}{\partial x}$$, then there is an $$L$$-preserving diffeomorphism $$\phi:(a,b)\to(c,d)$$ between two bounded intervals $$(a,b),(c,d)\subset\mathbb{R}$$ if and only if either $$b-a=d-c=2\pi k$$ or else $$2\pi(k{-}1) < b-a, d-c <2\pi k$$ for some integer $$k\ge1$$.

Thus, this gives a complete answer in the case $$n=1$$ and the locally transitive case. The need for the hypothesis of local transitivity is shown by this example: Let $$\phi:\mathbb{R}\to [0,1)$$ be a smooth function such that $$\phi(x)>0$$ when $$|x|<1$$ and $$\phi(x)=0$$ when $$|x|\ge 1$$. Let $$u:(-1,1)\to \mathbb{R}$$ satisfy $$u'(x) = 1/\phi(x)$$ when $$|x|<1$$. Then $$u:(-1,1)\to\mathbb{R}$$ is a diffeomorphism. Let $$L\subset\mathfrak{X}\bigl((-1,1)\bigr)$$ be the Lie algebra spanned by the vector fields $$\phi(x)\,\frac{\partial}{\partial x},\quad \cos\bigl(u(x)\bigr)\,\phi(x)\,\frac{\partial}{\partial x},\quad \sin\bigl(u(x)\bigr)\,\phi(x)\,\frac{\partial}{\partial x}.$$ These vector fields extend smoothly to the entire real line as zero vector fields where $$|x|\ge1$$, but one could also extend them smoothly to be periodic of period $$2$$. In either case, they vanish to infinite order at the odd integers. Clearly, there are infinitely many globally inequivalent ways to embed $$\mathfrak{sl}_2$$ as a subalgebra of $$\mathfrak{X}(\mathbb{R})$$ if one does not assume local transitivity. Classification is pretty hopeless.

Now assume that $$n>1$$. Here, the situation is somewhat more straightforward. The first thing to notice is the classical result that $$L=\mathfrak{sl}_{n+1}=\mathfrak{sl}_{n+1}(\mathbb{R})$$ has no proper subalgebras of codimension less than $$n$$. (This was known to Lie, at least in the complex case, i.e., for $$\mathfrak{sl}_{n+1}(\mathbb{C})$$, and the result for the real case follows immediately from this by complexification.) In fact, there is a complete descriptions of the set $$S_n$$ of subalgebras of $$L$$ that have codimension $$n$$. This is an $$n$$-manifold with two components $$S_n = \mathrm{SL}(n{+}1,\mathbb{R})/P_1 \cup \mathrm{SL}(n{+}1,\mathbb{R})/P_2\simeq \mathbb{RP}^n\cup \mathbb{RP}^n\,$$ where $$P_1$$ is the maximal parabolic consisting of the elements of $$\mathrm{SL}(n{+}1,\mathbb{R})$$ that preserve a given $$1$$-dimensional subspace of $$\mathbb{R}^{n+1}$$ and $$P_2$$ is the maximal parabolic consisting of the elements of $$\mathrm{SL}(n{+}1,\mathbb{R})$$ that preserve a given $$n$$-dimensional subspace of $$\mathbb{R}^{n+1}$$. (These two subgroups are not conjugate in $$\mathrm{SL}(n{+}1,\mathbb{R})$$, but there is an outer automorphism ($$a\mapsto (a^{-1})^\mathsf{T}$$) that carries one to the other. (Of course, when $$n=1$$, the space $$S_1=\mathbb{RP}^1\simeq S^1$$ has only one component.)

Now, suppose that $$L$$ is embedded as a subalgebra of $$\mathfrak{X}(M^n)$$, where $$M^n$$ is a smooth manifold of dimension $$n$$. Then, for any $$p\in M$$, the subalgebra $$L^0_p\subset L$$ consisting of the vector fields that vanish at $$p$$ has codimension at most $$n$$, so either $$L^0_p = L$$, in which case, all the vector fields in $$L$$ vanish at $$p$$, or else the codimension of $$L^0_p$$ in $$L$$ is exactly $$n$$. Let's remove the closed set of points where all the vector fields in $$L$$ vanish (which will not affect the completeness of any of the vector fields in $$L$$) and assume that $$L^0_p$$ always has codimension $$n$$.

Thus, we have a canonical map $$\phi:M\to S_n$$ given by $$\phi(p) = L^0_p$$.

Theorem: If $$L\subset\mathfrak{X}(M^n)$$ is a locally transitive subalgebra isomorphic to $$\mathfrak{sl}_{n+1}$$, then $$\phi:M\to S$$ is a smooth immersion (in particular, $$\phi$$ is a local diffeomorphism). Moreover, $$\phi_*$$ carries $$L$$ to the 'natural' copy of $$L$$ in $$\mathfrak{X}(S_n)$$ induced by the action of $$\mathrm{SL}(n{+}1,\mathbb{R})$$ on $$S_n$$.

The proof of this theorem is not hard, but it consists of a number of steps, whose details would take up a lot of space. The main points are these: First, let $$L^i_p\subset L$$ denote the subset of vector fields that vanish to order $$i{+}1$$ at $$p$$, then one first shows that $$L^i_p = 0$$ for $$i$$ suffciently large (a priori depending on $$p$$). Then, using the fact that $$L^0_p$$ contains a simple subalgebra isomorphic to $$\mathfrak{sl}_n$$, one uses its representation theory to show that this simple subalgebra cannot be contained in $$L^1_p$$ (which is a solvable ideal of $$L^0_p$$), and hence $$L^0_p/L^1_p$$ is an algebra containing $$\mathfrak{sl}_n$$. Using the nondegeneracy of the Killing form, one concludes that $$L^1_p$$ is dual, as a $$\mathfrak{sl}_n$$-representation, to $$L/L^0_p\simeq\mathbb{R}^n$$ and hence has dimension $$n$$. Thus, $$L^0_p/L^1_p$$ has dimension $$n^2$$ and must be isomorphic to $$\mathfrak{gl}_n$$. From this, it follows easily that $$L^2_p = (0)$$. Now, using these facts, it is easy to explicitly compute the differential of $$\phi$$ at $$p$$ and show that $$\phi'(p):T_pM\to T_{\phi(p)}S_n$$ is an isomorphism. Once one has the fact that $$\phi$$ is a local diffeomorphism, the final statement follows relatively easily.

Note that because $$S_n$$ has two components there are essentially two inequivalent global realizations of $$L=\mathfrak{sl}_{n+1}$$ as vector fields on $$\mathbb{RP}^n$$. By this, I mean that there are two Lie algebra homomorphisms $$\psi_i:L\to\mathfrak{X}(\mathbb{RP}^n)$$ such that there is no diffeomorphism $$u:\mathbb{RP}^n\to\mathbb{RP}^n$$ such that $$\psi_2(X) = u_*\bigl(\psi_1(X)\bigr)$$ for all $$X\in L$$. What is true instead is that there is an outer automorphism $$\tau:L\to L$$ such that $$\psi_2\bigl(\tau(X)\bigr) = u_*\bigl(\psi_1(X)\bigr)$$. This is, perhaps, a subtle point, but it shows that there really are two essentially different ways that $$L$$ can be realized as vector fields in dimension $$n$$. It is not clear which one should be called 'natural'.

Also, note that, if $$L\subset\mathfrak{X}(M)$$ consists entirely of complete vector fields, then $$\phi:M\to S_n$$ must be a covering space, in particular, if $$M$$ is connected, then it must be either $$S^n$$ or $$\mathbb{RP}^n$$. (Note that this is another place where $$n>1$$ differs from the case $$n=1$$.)

Remark: This non-uniqueness can be even more dramatic. A famous example is the split form of type $$\mathfrak{c}_2$$ (usually denoted as $$\mathfrak{sp}_4(\mathbb{R})$$), a Lie algebra of dimension $$10$$ that has two non-conjugate families of subalgebras of minimal codimension $$3$$, leading to two inequivalent ways that it can appear as a Lie algebra of vector fields on a $$3$$-manifold. One is as contact vector fields on $$\mathbb{RP}^3$$, and the other is as the conformal vector fields on $$\mathcal{N}^{2,1}$$, the manifold of null subspaces of dimension $$1$$ in Minkowski $$5$$-space, $$\mathbb{M}^{4,1}$$. This is the basis of the classical Klein correspondence.

For more exotic examples, the split form of the exceptional Lie algebra $$\mathfrak{g}_2$$ (of dimension $$14$$) has two non-conjugate subalgebras of the minimal codimension $$5$$. Since $$\mathfrak{g}_2$$ has no outer automorphisms, they are not even equivalent up to isomorphisms of the algebra. Thus, as both Cartan and Engel realized in 1893, $$\mathfrak{g}_2$$ can appears as two essentially different subalgebras of the vector fields in $$\mathbb{R}^5$$. This also happens for the split form of the exceptional Lie algebra $$\mathfrak{f}_4$$, and, correspondingly, it can be realized as two essentially different subalgebras of the vector fields in $$\mathbb{R}^{15}$$.

• Wow... that's a lot to process! Since you basically wrote a (beautiful) paper in answer to my (stupid) question, I assume that there isn't yet out there a unique book/article containing all the results you have mentioned. Jun 14, 2021 at 10:12
• @GiovanniMoreno: Your question was not stupid. It's a natural question; it and similar questions were studied seriously by the likes of Lie, Killing, Engel, and Cartan. Basically, the question is a special case of the question of how many 'inequivalent' ways a given Lie algebra can be realized as a Lie algebra of vector fields on some smooth manifold. They realized right away that the essential part of the problem, at least in the locally transitive case, was to classify the subalgebras of a given Lie algebra, but that's only a start, and they usually assumed real-analyticity. Jun 14, 2021 at 13:49