# How do I show this relation between exterior product and the projection of a tensor product

I have troubles understanding this whole problem starting at the definition. We have defined the exterior product as follows:

If $\alpha = \pi (a) \in \bigwedge^pV$ and $\beta = \pi(b) \in \bigwedge^q V$, then $$\alpha \wedge \beta = \pi (a\otimes b)$$ Where $\pi: T(V) \rightarrow \bigwedge^\ast V=T(V)/I(V)$ is the quotient projection

I have to show that for finite dimensional $V$ and vectors $v_1, v_2$ it holds that $$v_1 \wedge v_2 = \pi (1/2(u_1 \otimes u_2 - u_2 \otimes u_1))$$

Am I right, by assuming that $v_i=\pi(u_i)$? How do I use this to show this?

It should be stated in the problem but I think you are right about $v_i = \pi (u_i)$.
Hint: observe that $\frac{1}{2}((u_1 + u_2) \otimes (u_1 + u_2))= \frac{1}{2}(u_1 \otimes u_2 + u_2 \otimes u_1) + \frac{1}{2}(u_1 \otimes u_1 + u_2 \otimes u_2)$ is projected by $\pi$ to $0$ on one hand (projection of LHS) and to $\frac{1}{2}(u_1 \otimes u_2 + u_2 \otimes u_1)$ on the other (projection of RHS) then just use linearity of $\pi$.
• Oh! that was too obvious! So I see the left and the right hand side. How am I supposed to use linearity of pi? I have to show that $\pi(u_1\otimes u_2) = \pi (1/2( u_1\otimes u_2 - u_2\otimes u_1))$ right? – Pablo Elias Apr 15 '15 at 10:26
• Got it! I have to start at $\pi(u_1\otimes u_2 - 1/2(u_1+ u_2)\otimes(u_1+u_2))$ Thanks! – Pablo Elias Apr 15 '15 at 10:34