# Exterior Algebra as quotient

Given a vector space $W$, I understand what the tensor algebra $T(W)$ is, and I understand that the exterior algebra $\bigwedge W$ is defined as $\bigwedge W := T(W)/N$ where $N$ is the two-sided algebra ideal generated by $\{v \otimes w + w \otimes v | v,w \in W\}$. However, some texts quotient out by the ideal generated by $\{w \otimes w| w \in W\}$ instead of $\{v \otimes w + w \otimes v | v,w \in W\}$. Can someone explain why taking a quotient of either of these things yields the same algebra?

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$$v \otimes w + w \otimes v = (v+w) \otimes (v+w) - v\otimes v - w\otimes w$$ so the latter ideal contains the first one. For field of characteristic not 2, the first one contains the latter one. So in $char \neq 2$ cases they are the same.

I believe that $v \wedge v = 0$ is a condition you always want, which cannot be deduced from your first definition. So in characteristic 2, only the latter one is right. That's probably why some texts define it that way.

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+1 Much more thorough answer than mine. – user38268 Jan 31 '13 at 4:09
This was very helpful. Thanks. – nigelvr Jan 31 '13 at 23:57

This is a variant of the polarization trick: $$v \otimes w + w \otimes v = (v + w) \otimes (v + w) - v \otimes v - w \otimes w$$

Let $I$ be your ideal generated by all $v \otimes w + w \otimes v$ and let $J$ be the ideal generated by all $w \otimes w$.

The polarization identity shows that $I$ is contained in $J$. Of course, you can go the other way around to get $J \subseteq I$ by setting $v = w$ in your ideal if you are able to divide by $2$. The ideals are not the same if the scalar field for your vector space $W$ has characteristic $2$. People who deal with characteristic $2$ usually use the ideal $J$ because of this.

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+1 I love the polarisation identity. – user38268 Jan 31 '13 at 4:10

Not sure if this is a rigorous explanation but an intuitive way to think of this: The idea of quotiening out by the ideal $\{ v \otimes w+ w \otimes v | v,w \in W\}$ is that we impose the relation $$v \wedge w = - w\wedge v$$ in the quotient where $v \wedge w$ is the image of $v \otimes w$ under the canonical map. Now what happens when we quotient out by $\{w \otimes w| w\in W\}$, do we impose the same relation? Yes we do because

$$\begin{eqnarray*} 0 &=& (v +w) \wedge (w +v)\\ &=&(v \wedge w ) + (w \wedge v).\end{eqnarray*}$$

This should give you an idea why the quotients are isomorphic.

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