Take the 2-minute tour ×
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.

I am working on old qualifying problems involving tensor products. I am stuck on a statement about invertible elements in an exterior algebra and was wondering if this was a well known fact in a book somewhere. I think most of this notation is standard from dummite and foote but the notes I have been using are slightly different than what I have seen in textbooks so far.

Let $V$ be a finite dimensional vector space over a field $F$.

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

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

Let $x \in \wedge V$. So that $x = \sum_{k\geq 0} x_k$ where each $x \in \wedge^k V = T^k(V)/A^k(V)$

How do you prove that $x$ is invertible if and only if $x_0 \neq 0$

share|improve this question
Pardon my not knowing anything really about exterior algebras, but what does it mean for an $x$ to be invertible? Anticommutativity precludes an identity element wrt the wedge product, so it must mean something else.. –  anon Oct 19 '11 at 23:20
@anon: T^0(V) is just a copy of the field, so F ≤ ⋀V. Maybe the keyword is "graded commutative". (v∧w)∧x = x∧(v∧w) since that is two "swaps", (v∧w)∧x = −v∧x∧w = −−x∧v∧w. –  Jack Schmidt Oct 19 '11 at 23:35
add comment

1 Answer

up vote 2 down vote accepted

The set $I = \oplus_{k=1}^\infty T^k(V)$ is an ideal of $T(V)$ containing $A(V)$. I believe $\bigwedge V$ is a finite dimensional local algebra with unique maximal ideal $I/A(V)$. In particular, every element of $I/A(V)$ is nilpotent. In fact if $\dim(V) = n$, then $x^{n+1} = 0$ for every $x \in I$. In particular, the geometric series $\frac{1}{1-x} = \sum_{k=0}^\infty x^k$ converges (is a finite sum plus a bunch of 0s) for every $x \in I$, and shows that such elements are invertible. Multiplying by a nonzero element of the field, gives the result.

share|improve this answer
add comment

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.