Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The title says it all. For the uninitiated: Any map $f:V \to W$ induces a map $\wedge^k V \to \wedge^k W$ by $v_1 \wedge \cdots \wedge v_k \mapsto f(v_1)\wedge \cdots \wedge f(v_k)$, so $\wedge^k(-)$ is a functor from vector spaces to itself.

I have a map $\varphi:\wedge^k V\to \wedge^kV$ and I have reason to suspect that this map does not come fram a map $\psi:V \to V$.

I'm not sure how to prove this. It seems that any map in the image of $\wedge^k(-)$ must satisfy some kind of Plücker relations (similar to those of the Grassmannian), since if $\varphi \in \hom(V,V)$ is represented by the matrix $(x_{ij})$, then its image in $\hom(\wedge^k V,\wedge^k V)$ has matrix $(\det_{IJ}(x_{ij}))$, where $\det_{IJ}$ means take the determinant of the submatrix with indices $I$ and $J$.

I tried asking Macaulay2 to compute the ideal of relations, but even in the case $\dim V=4$ and $k=2$, it does not seem to be a feasible computation:

R = QQ[x_1..x_16]
M = genericMatrix(R,4,4)
I = minors(2,M)
numgens I
>> 36
S = QQ[y_1..y_36]
f = map(R,S,gens I)
ker f
>> ......???? <- to much for M2!

Is there some strategy to determine if my map comes from $\hom(V,V)$ other than computing these relations? If not, are the relations between the $k\times k$-minors of a $n\times n$-matrix known?


share|cite|improve this question
Interesting question. In your matrix description $I,J$ have cardinality $k$, right? And there is no harm in replacing $\hom(V,V)$ by $\hom(V,W)$, right? – Martin Brandenburg Mar 15 '13 at 14:37
@Martin: Yes, you're right. $I,J$ runs through all subsets cardinality $k$. – Fredrik Meyer Mar 15 '13 at 14:39
If $V$ has dimension $d$ and $k=d-1$, then there is a canonical(!) isomorphism $\wedge^{d-1} V \cong V^* \otimes \wedge^d V$, and $\wedge^d V$ is invertible. In particular we get a map $\wedge^{d-1} : \mathrm{End}(V) \to \mathrm{End}(V^*)$. It identifies with $f \mapsto f^*$ and is therefore an isomorphism. So interesting things only happen for $1<k<d-1$. – Martin Brandenburg Mar 15 '13 at 14:48
if $k \le \frac{1}{2}\dim V$ then you can try computing things like $\phi(v_1 \wedge v_{i_1} \wedge \cdots \wedge v_{i_{k-1}}) \wedge \phi(v_1 \wedge v_{j_1} \wedge \cdots \wedge v_{j_{k-1}})$. These have to be zero if $\phi$ comes from a map $V \to V$. – Eric O. Korman Mar 18 '13 at 12:59
up vote 2 down vote accepted

The (Zariski closure of the) image of your map is studied in the following paper of Bruns and Conca:

They stratify this closure by two numerical invariants rank and "small rank" and show that they correspond to $GL(V) \times GL(W)$ orbits. I never read the paper carefully, so I don't know if it will be of any use, but hopefully there are some interesting things in there for you.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.