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Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$. Set $$H=\{B\in\text{GL}(\bigwedge^k V) \, | \, B \,\text{ preserves pure tensors }\}$$

(i.e. $B \in H$ if it maps decomposable elements to decomposable elements).

$H$ is a closed subgroup of the Lie group $\text{GL}(\bigwedge^k V) $, hence it is an embedded Lie subgroup of it. Can we characterize it explicitly? What is $\dim H$? How does its Lie algebra look like?

Here is a lower bound: The exterior power map $\psi:\text{GL}( V) \to \text{GL}(\bigwedge^k V)$ defined by $\psi(A)=\bigwedge^k A$ is a smooth immersion*. Since $\text{Image}(\psi)\subseteq H$, $\dim H \ge \dim \text{GL}( V) =d^2$.

I think that this might be related to the following question: Let $\omega^I$ be a basis for $\bigwedge^kV$, whose elements are all decomposable. Is $\omega^I$ "induced" by a standard basis, i.e. $\omega^{i_1,\ldots,i_k}=v^{i_1} \wedge \ldots \wedge v^{i_k}$ for some basis $v_i$?

*One way to see that the exterior power map $\psi$ is an immersion, is that it is a locally-injective homomorphism of Lie groups, and in particular, it has a constant rank. The local injectivity (which comes from the fact that $\bigwedge^k A=\bigwedge^k B \Rightarrow A= \pm B$) then implies that $\psi$ is an immersion, due to the constant rank theorem.

Asaf Shachar
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  • Do you have examples of $B \in H$ not of the form $\bigwedge^k A$ for any $A$ ? – Maxime Ramzi Jan 20 '19 at 15:52
  • I'm not sure I understand; there are elements of $\operatorname{GL}(\bigwedge^kV)$ that preserve decomposable elements, but that are not induced by automorphisms of $V$. In the simplest case, when $d=4$ and $k=2$, the set of decomposable elements is the zero locus of $$x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23},$$ with respect to the standard basis $e_i\wedge e_j$ of $\bigwedge^2V$. The element of $\operatorname{GL}(\bigwedge^kV)$ that switches $x_{12}$ and $x_{34}$ and fixes the other $x_{ij}$, for example, clearly stabilizes the Grassmannian, but is not induced by an automorphism of $V$. – Servaes Jan 22 '19 at 15:27
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    @Servaes : I don't understand everything in the given link; but it does state that there are exceptions ("except when the dual map is an automorphism - when $n=2p$"; here $n$ is $d$ and your $d$ is $4=2\times 2$ so that might be it) – Maxime Ramzi Jan 22 '19 at 16:04
  • Agreed, problem solved! – Servaes Jan 22 '19 at 16:31
  • @Servaes Well, at least the problem is solved when $V$ is a complex vector space. I am interested in the real case though. Maybe we can reduce the real case to the complex case, but I am not sure. (You can see here.). – Asaf Shachar Jan 23 '19 at 12:45

1 Answers1

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Here is a solution for the case where $V$ is a complex vector space. (I guess that the real case can be reduced to the complex one, see some subtle points below). It turns out that $H$ is pretty restricted: When $d \neq 2k$, $H=\text{Image}(\psi)$, i.e. every map which preserves pure tensors is an exterior power of some $A \in \text{GL}$.

When $d=2k$, we have additional automorphisms, of the form $\star \circ \bigwedge^k A$, where $\star:\bigwedge^k V \to \bigwedge^k V$ is the Hodge dual operator w.r.t some fixed metric on $V$.

For a reference, see here.

Here are some natural questions which arise from this representation claim:

1. Is this representation redundant in some way?

i.e. if $\star$ equals to some exterior power, then it is not a "true" addition to the subgroup $H$. However, the Hodge is never an exterior power: Indeed, suppose that $\star=\bigwedge^k A$ for some $A \in \text{GL}(V)$. Let $v_1,\dots,v_d$ be an orthonormal basis for $V$.

$\text{span}(Av_1,\dots,Av_k)=\text{span}(v_{k+1},\dots,v_d)$. Switching $v_k$ and $v_{k+1}$, we also deduce that $\text{span}(Av_1,\dots,Av_{k-1},Av_{k+1})=\text{span}(v_k,v_{k+2},\dots,v_d)$, so $$\text{span}(Av_1,\dots,Av_{k-1})= \text{span}(v_{k+2},\dots,v_d).$$

In particular, we have proved that for any orthonormal basis $v_1,\dots,v_{k-1},v_{k},v_{k+1},\dots,v_d$, $$\text{span}(Av_1,\dots,Av_{k-1}) \subseteq \text{span}(v_{k+2},\dots,v_d). \tag{1} $$ Now, switching between $v_{k+1}$ and $v_{k+2}$, this also implies
$$\text{span}(Av_1,\dots,Av_{k-1}) \subseteq \text{span}(v_{k+1},v_{k+3},v_{k+4},\dots,v_d) \tag{2}.$$

Combining $(1)$ and $(2)$ together, we deduce that $$\text{span}(Av_1,\dots,Av_{k-1}) \subseteq \text{span}(v_{k+3},\dots,v_d) $$, which implies that $A$ maps the $k-1$ dimensional space $\text{span}(v_1,\dots,v_{k-1})$ to the $k-2$ dimensional space $\text{span}(v_{k+3},\dots,v_d) $, i.e. $A$ is not invertible. This implies $\bigwedge^k A$ is also singular, so it cannot be equal to a Hodge dual. Contradiction.

2. What about automorphism of the form $ \bigwedge^k A \circ \star $? or $ \star \bigwedge^k A \circ \star $?

Well, using the equality $\star \circ \bigwedge^k A=\det A \bigwedge^k A^{-T} \circ \star=\bigwedge^k (\sqrt[k]{\det A}\cdot A^{-T} )\circ \star$ we can "move" $\star$ from one side of $\bigwedge^k A$ to the other one. (Here we used the assumption that $V$ is a complex vector space, since otherwise $\sqrt[k]{\det A}$ does not exist if $k$ is even, and $\det A<0$). Maybe in that case we need to allow multiplication of exterior powers by negative scalars).

3. How do we account for all the different Hodge duals of all the different metrics, by using the Hodge operator induced by a single metric?

Since the map $g \to \star_g$ is smooth and injective up to a conformal rescaling of the metric (that is $\star_g=\star_h$ if and only if $g=ch$ ), one expects to obtains a continuous family of different Hodge duals of dimension $\dim(\text{Psym_d})-1$, so the dimension of $H$ should increase accordingly. However, this is not really what happens, since we can or express every Hodge dual in terms of one specific representative, by compensate for the difference via some exterior power of a mediating map.

(Actually, I think that we can induce a Hodge dual using any non-degenerate bilinear form, symmetric or not). Here are the details:

Suppose $g,h$ are two metrics on $V$. Then we can write $g(v,w)=h(v,Tw)$ for some $T \in \text{GL}(V)$. It is obvious that $T$ is $h$-positive, and $h$-symmetric, hence there exist a uniqe $\sqrt T$ which is also $h$-positive and symmetric. Thus, we have $g(v,w)=h(\sqrt Tv,\sqrt Tw)$, i.e. $\sqrt T:(V,g) \to (V,h)$ is an orientation-preserving isometry, hence it commutes with the Hodge dual, i.e.

$\bigwedge^k \sqrt T \circ \star_g= \star_h \circ \bigwedge^k \sqrt T$. So, now we can write

$$ \star_g \circ \bigwedge^k A=\bigwedge^k \sqrt T^{-1} \circ \star_h \circ \bigwedge^k \sqrt T A,$$

and since we only know how to move $\star_h$ from one side of an exterior power to the other side (see point $2$), this representation can be reduced to $\star_h \circ \bigwedge^B$ for some $B \in \text{GL}(V)$.

Similarly, we can represent creatures of the form of $\star_h \circ \star_g$...by expressing $\star_g$ via $\star_h$ or vice-versa.

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Asaf Shachar
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