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$\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\Cof}{\operatorname{Cof}}$ $\newcommand{\Det}{\operatorname{Det}}$ $\newcommand{\id}{\operatorname{Id}}$

Let $V$ and $W$ be $d$-dimensional, oriented inner-product spaces, and let $A\in\Hom(V,W)$ be an orientation-preserving map. Suppose that $A$ satisfies $$ \bigwedge^{d-k} A \circ \star_V^k= \star_W^{k} \circ \bigwedge^k A \neq 0, \tag{1} $$

for a single $1 \le k \le d-1$.

Question: Is it true that $A$ is conformal? If not, can we characterise the maps which satisfy this?

Comment (1): Condition $(1)$ implies $A$ is invertible. (Proof at the end). If we omit the part $\bigwedge^k A \neq 0$, then every map with rank less than $\min(k,d-k)$ would satisfy the commutation property (and won't be conformal).

Comment (2): Condition $(1)$ is symmtric in $k,d-k$.

Comment (3): I guess that if $k \neq d-k$, then $A$ needs to be an isometry. (The heuristics is this: Condition $(1)$ says the action on $A$ on parallelepipeds of dimensions $k,d-k$ is "the same", and when the scales are different, this is a rigidity constraint).

Edit(1): I proved that if $A$ is conformal, and $k \neq d-k$, then $A$ is an isometry, so we are back at the conformality question.

Partial progress (reformulation and proof for $k=1,d-1$):

Define $\Cof^k A := (-1)^{k(d-k)} \star_W^{d-k} \circ \bigwedge^{d-k} A \circ \star_V^k$. Note that $\Cof^k A \in\Hom( \Lambda_k(V) ,\Lambda_k(W))$. It can be proved that

$$\Det A \cdot \id_{\Lambda_k(V)} = \bigwedge^k A^T \circ \Cof^k A \tag{2}$$

for any map $A \in \Hom(V,W)$.

Condition $(1)$ implies that $\Cof^k A = \bigwedge^k A$. Plugging this into equation $(2)$, we get

$$ \Det A \cdot \id_{\Lambda_k(V)} = \bigwedge^k A^T \circ \bigwedge^k A =\bigwedge^k (A^T \circ A) \tag{3}.$$

Denote $A^TA=S \in \Hom(V,V)$. Then we obtained

$$ \bigwedge^k (\Det S)^{\frac{1}{2k}}\id_V =\sqrt{\Det S} \cdot \id_{\Lambda_k(V)} =\bigwedge^k S \tag{4}.$$

Does this imply $S=(\Det S)^{\frac{1}{2k}}\cdot \id_V$?

If the wedge was "injective" in the sense that $\bigwedge^k S=\bigwedge^k T \Rightarrow S=T,$ we were done. Of course, this injectivity does not hold in general, since for $S=0$ any map $T$ of rank smaller than $k$ would satisfy $\bigwedge^k T=0$.

For $k=1$ the injectivity holds, so we are done. (We get $A^TA=\det A \id_V$. If $\det A= 0$ we get $A^TA=0 \Rightarrow A=0$ which contradicts the assumption. So $\det A \neq 0$, and $A$ is conformal).

Since condition $(1)$ is symmetric in $k,d-k$, the answer is also positive for $k=d-1$


Proof that condition $(1)$ implies invertibility:

Suppose by contradiction $\ker A \neq 0$, and let $r=\operatorname{rank}(A)=\dim\big((\ker A)^{\perp}\big)$. Since $ \bigwedge^k A \neq 0 \Rightarrow \operatorname{rank}(A) \ge k$, condition $(1)$ implies $r \ge k$. Let $v_1,\dots,v_r$ be an orthonormal basis for $(\ker A)^{\perp}$, and let $w_1,\dots,w_{d-r}$ be an orthonormal basis for $\ker A$.

Then $$\bigwedge^{d-k} A \circ \star_V^k (v_1 \wedge \dots \wedge v_k)= \star_W^{k} \circ \bigwedge^k A (v_1 \wedge \dots \wedge v_k)$$

Since $A|_{(\ker A)^{\perp}}$ is injective, the RHS is non-zero, while the LHS equals $\pm \bigwedge^{d-k} A (v_{k+1} \wedge \dots \wedge v_r \wedge w_1 \dots \wedge w_{d-r})=0 $, a contradiction.

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  • 1
    $\begingroup$ I suppose you assume that $\wedge^k A \not= 0$? $\endgroup$ – Gunnar Þór Magnússon May 31 '17 at 9:03
  • $\begingroup$ You are right. Actually I am not sure that even this assumption suffices. For instance suppose $d=10$, $k=3$, than isn't it possible to have a map $A$ of rank $8$ (so it's not conformal, since it's not invertible) which satisfies the condition for $k=3$? I am even ready to assume $A$ is invertible. (Although it is interesting to see what happens when we only assume $\wedge^k A \neq 0$). $\endgroup$ – Asaf Shachar May 31 '17 at 12:34
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    $\begingroup$ I do not think your rank $8$ map will work. You could pick an orthogonal basis $v_1,..,v_3,w_1,...,w_7$ such that the $Av_i$ are linearly independent, yet the $Aw_i$ are not. For example by choosing $w_7$ out of the Nullspace of $A$. Then $\star \wedge^3 Av_i \neq 0$, yet $\wedge^7 A\star w_i =0$. $\endgroup$ – mlk May 31 '17 at 13:28
  • $\begingroup$ @mlk Thanks, you are right. I think I can prove (using your idea) that $\wedge^k A \neq 0$ (together with the commutation condition in the question) implies $A$ must be invertible. I will update the question to mention this (and provide the proof). $\endgroup$ – Asaf Shachar May 31 '17 at 14:24
  • $\begingroup$ Asaf: I do not have time to think about this but everything reduces to the case when $V=W$ and $A$ is a diagonal matrix (just use the SVD). Then $\Lambda^i A$ is again diagonal (with respect to the standard basis on the exterior power). Now, you should be able to settle the question one way or another by an explicit computation since commuting with the Hodge star in the diagonal case is visible by looking at the standard basis in the exterior power. $\endgroup$ – Moishe Kohan May 31 '17 at 16:34
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Here is a proof for "diagonal maps" $A:V \to V$:

(The general case can probably be reduced to this case using SVD):

Let $v_1,...,v_d$ be a positive orthonormal basis for $V$, and suppose $Av_i=\sigma_iv_i, \sigma_i \ge 0$. By the remark in the question, we can assume $A$ is invertible so all $\sigma_i > 0$.

Then

$$\bigwedge^{d-k} A \circ \star_V^k (v_{i_1} \wedge \dots \wedge v_{i_k})=\bigwedge^{d-k} A (v_{i_{k+1}} \wedge \dots \wedge v_{i_d})= Av_{i_{k+1}} \wedge \dots \wedge Av_{i_d}=$$

$$\sigma_{i_{k+1}} v_{i_{k+1}} \wedge \dots \wedge \sigma_{i_d} v_{i_d}=\Pi_{j=k+1}^d \sigma_{i_j} \cdot (v_{i_{k+1}} \wedge \dots \wedge v_{i_d}).$$

On the other hand,

$$ \star_V^{k} \circ \bigwedge^k A (v_{i_1} \wedge \dots \wedge v_{i_k})=\star_V^{k} \big( \Pi_{j=1}^k \sigma_{i_j} \cdot (v_{i_1} \wedge \dots \wedge v_{i_k}) \big)= \Pi_{j=1}^k \sigma_{i_j}(v_{i_{k+1}} \wedge \dots \wedge v_{i_d}).$$

Now, the condition $\bigwedge^{d-k} A \circ \star_V^k= \star_V^{k} \circ \bigwedge^k A $ implies $$\Pi_{j=1}^k \sigma_{i_j}=\Pi_{j=k+1}^d \sigma_{i_j}.$$

for any choice of indices $1 \le i_1,\dots,i_k \le d$. Write explicitly $$ \sigma_{i_1} \cdot \sigma_{i_2} \cdot \dots \cdot \sigma_{i_k}=\sigma_{i_{k+1}} \cdot \sigma_{i_{k+2}} \cdot \dots \cdot \sigma_{i_d}, \tag{1}$$ and switch $i_k$ and $i_{k+1}$, so

$$ \sigma_{i_1} \cdot \sigma_{i_2} \cdot \dots \cdot \sigma_{i_{k+1}}=\sigma_{i_k} \cdot \sigma_{i_{k+2}} \cdot \dots \cdot \sigma_{i_d} \tag{2}$$ also holds.

Dividing $(2)$ by $(1)$, we get $$ \frac{\sigma_{i_k}}{\sigma_{i_{k+1}}}=\frac{\sigma_{i_{k+1}}}{\sigma_{i_{k}}} \Rightarrow \sigma_{i_k}=\sigma_{i_{k+1}}.$$

Since $i_k,i_{k+1}$ were arbitrary indices, we conclude all the $\sigma_i$ are equals, so $A$ is conformal.

Note: The proof indeed "breaks" when $k=0,d$ since we cannot switch two indices between different sides of the equality. In that case the commutation property is equivalent to $\det A=1$.

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  • $\begingroup$ Very good. Now use the singular value decomposition $A=UDV$, $D$ is diagonal, $U, V$ orthogonal, plus the fact that isometries $U, V$ commute with Hodge star and that if $A, B$ commute with Hodge star, so does $AB$. This will conclude the proof. $\endgroup$ – Moishe Kohan Jun 1 '17 at 17:53
  • $\begingroup$ Yes, you are right. On a second thought, the proof becomes a bit simpler when using the reduction I made in the question and orthogonal diagonalization, instead of proving it directly using SVD (you can see the new proof below). Anyhow, it does not really matter, the question is solved:) $\endgroup$ – Asaf Shachar Jun 2 '17 at 15:12
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$\newcommand{\Hom}{\operatorname{Hom}}\newcommand{\Cof}{\operatorname{Cof}}\newcommand{\Det}{\operatorname{Det}}\newcommand{\id}{\operatorname{Id}}$Here is an "alternative" answer, which is a bit simpler than the original:

Instead of working (implicitly) with SVD, we note the main claim was reduced (in the question) to the following:

Let $S \in \Hom(V,V)$ be symmetric and positive-definite (i.e $S^T=S$, $S$ has positive eigenvalues). Suppose

$$ \bigwedge^k \id_V = \id_{\Lambda_k(V)} =\bigwedge^k S \tag{4}.$$

Then $S=\id_V$.


Comment: This claim does not hold if we relax the positivity requirement, i.e take $S=-\text{Id}$, $k$ even.

Proof:

By orthogonal diagonalization,we can assume $S$ is diagonal: $sv_i=\sigma_i v_i,\sigma_i > 0$, where $v_1,...,v_d$ is an orthonormal basis for $V$.

$$v_{i_1} \wedge \dots \wedge v_{i_k}=\bigwedge^{k} S (v_{i_1} \wedge \dots \wedge v_{i_k})=(\Pi_{j=1}^k \sigma_{i_j}) (v_{i_1} \wedge \dots \wedge v_{i_k}),$$ implies $ \Pi_{j=1}^k \sigma_{i_j}=1,$ for any choice of indices $1 \le i_1,\dots,i_k \le d$. Write explicitly $$ \sigma_{i_1} \cdot \sigma_{i_2} \cdot \dots \cdot \sigma_{i_{k-1}} \cdot \sigma_{i_k}=\sigma_{i_1} \cdot \sigma_{i_2} \cdot \dots \cdot \sigma_{i_{k-1}} \cdot \sigma_{i_{k+1}}$$

(we switched $i_k$ and $i_{k+1}$), we deduce $\sigma_{i_k}=\sigma_{i_{k+1}}.$

Since $i_k,i_{k+1}$ were arbitrary indices, we conclude all the $\sigma_i=\sigma$ are equals. Now $\sigma^k=\Pi_{j=1}^k \sigma_{i_j}=1$ implies $\sigma=1$, so $S=\id_V$.

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