# Skew-symmetry of wedge product when vector space $W$ fails into a direct sum $U\oplus V$.

I'm trying to compute some calculus which need the exterior algebra $\Lambda(W)$ when the vector space $W$ is the direct of two vector subspaces $U$ and $V$ (in fact $V$ and $V^*$ but I think this no matters here). I know the general decomposition

$$\Lambda^p(W)= \oplus_{q+r=p} \Lambda^q(U)\otimes\Lambda(V)$$

and also the decomposition of the total vector space $\Lambda(W)$:

$$\Lambda(V)=\Lambda(U)\otimes\Lambda(V).$$

I have decided to start studying the simple case $X\wedge Y$ for $X,Y\in W$ and the symmetry $(-1)^? Y\wedge X$. As a vector space, $X\wedge Y = -Y\wedge X$ clearly. On the other hand, if $\pi_U:W\rightarrow U$ and $\pi_V:W\rightarrow V$ are the projections, then

$$X\wedge Y = (\pi_U(X+\pi_V(X))\wedge((\pi_U(Y+\pi_V(Y)) =\\ \bigg(\pi_U(X)\wedge\pi_U(Y)\bigg) + \bigg(\pi_V(X)\wedge\pi_V(Y)\bigg) + \bigg(\pi_U(X)\otimes\pi_V(Y) + \pi_V(X)\otimes \pi_U(Y)\bigg)$$

Clearly $\pi_U(X)\wedge\pi_U(Y) = - \pi_U(Y) \wedge \pi_U(X$), and same works for $V$. But in $U\times V$ we don't have this change of sign. Besied, if I compute $Y\wedge X$ in the same way, I would have one term of the form

$$\pi_U(Y)\otimes \pi_V(X) + \pi_V(Y)\otimes\pi_U(X),$$ different from the above. At this point, the unique solution I can think is that the change of signs occurs in the $\Lambda^q(U)$ and $\Lambda^r(V)$ subspaces whereas in the tensorial product $\Lambda^q(U) \otimes \Lambda^r(V)$ we use the canonical isomorphism $A\otimes B \cong B\otimes A$ for any vectorial spaces (or $R$-modules) $A$ and $B$.

1- Am I right?

2- Does it matter that in my concrete case $W=V\oplus V^*$ for a vector space $V$?

PD: I am always considering finite dimensional vector spaces.

$$\Lambda(W) \cong \Lambda(U) \otimes \Lambda(V)$$
is an isomorphism of graded vector spaces with the obvious notion of tensor product; it also holds as an isomorphism of graded algebras, but only with the Koszul sign rule applied to the tensor product on the right. This means that if $a, b \in \Lambda(U)$ and $c, d \in \Lambda(V)$ are homogeneous then
$$(a \otimes c) (b \otimes d) = (-1)^{|b| |c|} (ab \otimes cd)$$
owing to the need to switch $c$ and $b$ to write down this product. This should correct the calculation.
• You answer focus on the other part of my problem: working with homogeneous elements not in $W$ but in $U$ and $V$. It is also useful form, but I was worried for homogeneous elements in $W$. I mean, $a\otimes b$ and $c\otimes d$ don't represent homogeneous elements $X$ and $Y$ in $\Lambda(W)$ and then your reply doesn't answer completely my question (or so I think). – Dog_69 Dec 3 '17 at 22:17