2
$\begingroup$

Let $V$ be a one dimensional graded $\mathbb Q$-vector space; $V=\bigoplus_{i\geq 0}V_i$ and all $V_i$ are zero except $V_{2n+1}$ for some given $n$. Let $v$ be a generator of $V_{2n+1}$. Now take the free commutative graded algebra $\Lambda V$ on $V$. I want to understand the grading on $\Lambda V$. I know that $\Lambda V$ is graded as $\Lambda V=\bigoplus_{i\geq 0}{\Lambda^iV}$ where each $\Lambda^iV$ has a basis consisting of all possible products of $i$ elements from the basis of $V$. In our case the only basis element is $v$ and $v^2=0$ so the only product is $v$ of lenght 1, hence $\Lambda V=\Lambda^0V\oplus \Lambda^1V$ where $\Lambda^0V$ is isomorphic to $\mathbb Q$ with basis $1$ and $\Lambda^1V$ is isomorphic to $V$ with basis $v$. Is this correct? is it possible to have the following grading : $\Lambda V=\Lambda^0V\oplus \Lambda^{2n+1}V$?

I think it is not possible because we can not have an element of length $2n+1$ out of $v$ since $v^2=0$. Thank you for your help!!

$\endgroup$
1
$\begingroup$

The decomposition of $\Lambda V$ you give is useful when considering the length, as you refer to it, or number of factors in a wedge product $x = v_1 \wedge \ldots \wedge v_k \in \Lambda V$.

To figure out the grading, consider that the free commutative graded algebra usually has a universal property, namely being universal for maps from a graded vector space into a graded commutative, graded $\mathbb{Q}$-algebra: Given such a map of graded modules $f:V \to U(E)$, where $U$ is the forgetful functor from graded algebras to graded vector spaces, there should be a unique map of graded algebras $\hat f: \Lambda V \to E$ extending $f$.

In particular, $V$ can be taken to be contained in $\Lambda V$ and the inclusion should respect the grading, hence, the grading you propose yourself with $\mathbb{Q}$ in degree $0$ and $\Lambda^1 V = V$ (!) in degree $2n+1$ is the way to go here.

In general, elements of the form $v \in V$ keep their grading, and products or elements of degree higher than one add their gradings: $$ \operatorname{deg} ( v_1 \wedge \ldots \wedge v_k) = \sum_{i=1}^k \operatorname{deg}(v_i). $$

The confusion probably stems from the fact that if $V$ is a vector space interpreted as a graded vector space sitting in degree $1$, then in $\Lambda V$ the notion of length and degree conincide.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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