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Let $V$ be a rational graded vector space. Denote by $\Lambda V$ the free graded commutative algebra on $V$. Suppose that $V$ is one dimensional with basis $\{e\}$, then I read that the structure of $\Lambda V$ depends on the degree of $e$ wether it is odd or it is even. I understand that commutiativity implies that $e^2=(-1)^{|e|^2}e^2$ and in particular when $|e|$ is odd then $e^2=0$ hence the exterior algebra structure. My question is what is exactly the degree $|e|$ of $e$ on which the result depends, does it refer to the grading on $V=\oplus_i V_i$ so that $e\in V_{|e|}$ or to the grading of $\Lambda V=\oplus_i\Lambda^iV$ so that $e$ is a product of $|e|$ elements. I'm confused because given a grading on $V=\bigoplus_{i\geq 0}V_i$ there many notions of grading on $\Lambda V$ that is there is the notion of $\Lambda^iV$ of words of length $i$, there is also the notion of $(\Lambda V)^i$ of words of degree $i$, there is also the notion of $\Lambda V_i$ where we take the free commutative graded algebra on $V_i$ for each $i$, hence we have three gradings :

$$\Lambda V=\bigoplus_{i\geq 0}\Lambda^iV$$ $$\Lambda V=\bigoplus_{i\geq 0}(\Lambda V)^i$$ $$\Lambda V=\bigoplus_{i\geq 0}\Lambda V_i$$

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Note that the degree of an element in $\Lambda ^n V$ is not in general $n$. The degree of $e$ is the same in $V$ and $\Lambda V$.

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  • $\begingroup$ How is that!! an element in $\Lambda^nV$ is by definition of degree $n$.. $\endgroup$ – palio Feb 12 '13 at 8:08
  • $\begingroup$ No, how do you grade $V\otimes W$ for example? The degree of $a\otimes b$ in $V\otimes W$ is $|a|+|b|$, right? The degree the of $a_1\wedge\cdots \wedge a_n$ is simply $|a_1|+\dots +|a_n|$. See en.wikipedia.org/wiki/… $\endgroup$ – bashar Feb 12 '13 at 9:40

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