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Let $V=k^n$ for an algebraically closed field $k$ of characteristic 0, and let $W \subseteq V$ a subspace. Let $G_W\subseteq GL(V)$ be the set of invertible linear maps that preserve $W$, i.e. $$ G_W=\{x \in GL(V): x(W)=W\}, $$ and let $$ \mathfrak{g}_W=\{X \in \mathfrak{gl}(V) : X(W) \subseteq W\}, $$ where $\mathfrak{gl}(V)$ is the Lie algebra of $GL(V)$, identified with the set of linear maps on $V$ under the commutator bracket. It is known (see e.g. Humphrey's Linear Algebraic Groups section 13.8) that $\mathfrak{g}_W$ is the Lie algebra of $G_W$.

My question is: If $Y \subseteq V$ is an arbitrary irreducible affine variety, what is the Lie algebra of its preserver $G_Y$? I have a guess that would be consistent with the above result:

Guess: Suppose $Y \subseteq V$ is a homogeneous irreducible affine variety (i.e. $Y$ is a cone: $v \in Y \iff \alpha v \in Y$ for all $\alpha \in k$). Let

$$ \mathfrak{g}_Y=\{X \in \mathfrak{gl}(V) : X(\mathscr{L}(Y)) \subseteq \mathscr{L}(Y)\}, $$ where $\mathscr{L}(Y) \subseteq V$ is the tangent space to $Y$ at $0$, i.e., if $Y=V(f_1,\dots, f_m)$, then $\mathscr{L}(Y)=V(d_0f_1,\dots, d_0 f_m)$, where $d_0 f (x)= \sum_{j=1}^n \frac{\delta f}{\delta x_i} (0) x_i$. Then $\mathfrak{g}_Y$ is the Lie algebra of $G_Y$.

Is this guess correct? If so, I would appreciate a proof or a reference to a proof.

A quick proof of the above result for when $Y=W$ is a subspace is provided by ShinyaSakai in the comments to this question

EDIT @abx quickly disproved my guess, so my new question is simply: What is the Lie algebra of $G_Y$? I would appreciate any relevant references in this vein.

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  • $\begingroup$ Wrong. If $Y$ is not contained in a hyperplane, the $f_i$ are homogeneous of degree $>1$, hence their derivatives vanish at $0$, and $\mathfrak{g}_Y=\mathfrak{gl}(V)$. $\endgroup$
    – abx
    Commented Mar 22, 2020 at 20:57
  • $\begingroup$ @abx Thanks! Do you have any idea what the Lie algebra of $G_Y$ is then (or references to results in this vein)? $\endgroup$
    – Ben
    Commented Mar 22, 2020 at 21:08
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    $\begingroup$ See mathoverflow.net/questions/165900/… for the cubic case. I believe if the degree is > 2 the group is usually finite, so you won't get interesting information from the Lie algebra. (Of course for degree 2 you get the orthogonal group) $\endgroup$
    – Kevin Casto
    Commented Mar 23, 2020 at 3:34

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Doc, you are right but only amorally. You need to replace the tangent vectors with jets to capture the behavior of your cone.

Let $I(Y)$ be the ideal of zeroes of your $Y$. Then $$ Lie (G_Y) = \{ X \in {\mathfrak{gl}}(V) | X(I(Y))\subseteq I(Y)\}. $$ Now you know that $I(Y)$ is homogeneous. Pick a finite set of its generators. Let $n$ be the highest degree of a generator from your set. Consider $n$-cojets $$ J^\ast(Y) := I(Y)/(I(Y)\cap I(0)^{n+1}) \subseteq J^\ast := I(0)/I(0)^{n+1} $$ and $n$-jets $$ J(Y) := J^{\ast}(Y)^\perp \subseteq J := (J^\ast)^\ast $$ where $I(0)$ is the principal ideal. This yields a desired "finite-dimensional" condition $$ Lie (G_Y) = \{ X \in {\mathfrak{gl}}(V) | X(J(Y))\subseteq J(Y)\}. $$

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  • $\begingroup$ I'm having trouble verifying your first line $Lie(G_Y)=...$ How does $X \in \mathfrak{g}\mathfrak{l}(V)$ act on $f \in I(Y)$? $\endgroup$
    – Ben
    Commented May 26, 2020 at 21:25
  • $\begingroup$ $f\in k[V]$ on which $X$ naturally acts $\endgroup$
    – Bugs Bunny
    Commented May 27, 2020 at 6:28
  • $\begingroup$ Okay, we can think of the action of $GL(V)$ on $V$ as an algebraic group homomorphism $GL(V) \rightarrow GL(k[V])$, the differential of which is a Lie algebra homomorphism $\mathfrak{g}\mathfrak{l}(V) \rightarrow \mathfrak{g}\mathfrak{l}(k[V])$, which gives our action of $\mathfrak{g}\mathfrak{l}(V)$ on $k[V]$. Is there a formula for this action? I would greatly appreciate a reference. $\endgroup$
    – Ben
    Commented May 27, 2020 at 14:03
  • $\begingroup$ Just come from an algebraic cloud down to the analytic Earth. The elementary matrix $E_{i,j}$ acts as a vector field $x_i\frac{\partial}{\partial x_j}$. I am not sure about a reference. Probably, Theorie der Transformationsgruppen by Lie :-)) $\endgroup$
    – Bugs Bunny
    Commented May 28, 2020 at 19:20
  • $\begingroup$ Thanks. I think I see it up here in the clouds. Any algebraic group $G$ action on an affine variety $X$ induces a natural action of $\mathfrak{g}$ on $k[X]$. A standard example is when $X=G$ and $g \cdot h= hg^{-1}$, in which case $\mathfrak{g}$ acts on $k[G]$ by right convolution (i.e., as a left-invariant derivation). A good reference is Humphreys Linear Algebraic Groups. It's not hard to extend this argument to get a formula for an arbitrary variety $X$. $\endgroup$
    – Ben
    Commented May 30, 2020 at 17:08

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