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 following question is related to this post Showing an isomorphism between exterior products of a free module about the existence of a canonical isomorphism of the exterior product of free modules and in particular details the pathological behavior of certain isomorphisms given conditions on the rank of the module.

Let $F$ be a free $R$-module of rank $ n\geq 3$ where $R$ is a commutative ring with identity.

Let $T(F) = \oplus_{k=0}^{\infty} T^{k}(F)$ where $T^k(F) = F \otimes F \otimes \ldots \otimes F$ is tensor product of $k$ modules.

Let $\wedge F $ denote the exterior algebra of the $R$-module $F$, that is the quotient of the tensor algebra $T(F)$ by the ideal $A(F)$ generated by all $x \otimes x$ for $x \in F$.

Suppose $1\leq p \leq n-1$ and that $n \neq 2p$

How do we show there cannot exist a bijective module homomorphism $f: \wedge^p F \rightarrow \wedge^{n-p} F$ such that $f \circ (\wedge^p \psi) = (\wedge^{n-p} \psi ) \circ f $ for all $\psi \in Aut(F)$?

share|cite|improve this question
1  
I think that you mean "$f\circ (\wedge^p\psi) = (\wedge^{n-p} \psi)\circ f$" in you last line. – Matt E Oct 21 '11 at 4:31
    
thank you I just made the correction. – user7980 Oct 21 '11 at 4:54
    
The last sentence is ungrammatical -- do you mean "such that" (without "if")? – joriki Oct 21 '11 at 7:02
    
What's the counterexample? I can't find one in the other question you link to. – joriki Oct 21 '11 at 7:07
1  
You should first answer by yourself a similar but simpler problem: prove that we can't find an isomorphism $f:F \to F^\vee$ (the dual of $F$) such that $f \circ \psi = (\psi^\vee) \circ f$ for every $\psi \in GL(F)$. Hint: chose a basis write things down in terms of matrices. – AFK Oct 22 '11 at 21:25

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.