Let's say I have two vector spaces $V,W$ , and we have the graded algebras $\Lambda(V),\Lambda(W)$, each with an operation $\wedge$. I'd like to know if there are "many" $\wedge$ operators, or if there is just one, in the following sense. I can construct tensor products $\Lambda^p(V)\otimes\Lambda^q(W)$. I imagine that I could act with a "V wedge" which takes


$ \wedge\_V : (\Lambda^p(V)\otimes\Lambda^q(W))\times (\Lambda^r(V)\otimes\Lambda^s(W))\to \Lambda^{p+r}(V)\otimes\Lambda^q(W)\otimes\Lambda^s(W)$


and similarly for $\wedge\_W$. I don't know if these are good operators to consider or if there is some reason it's silly. Now I could also imagine an "everything wedge" which takes


$\wedge\_{all} : (\Lambda^p(V)\otimes\Lambda^q(W))\times (\Lambda^r(V)\otimes\Lambda^s(W))\to \Lambda^{p+r}(V)\otimes\Lambda^{q+s}(W)$.


I again don't know enough algebra if this is a silly thing to consider or if it's some natural thing. When I am wedging, should I be specifying which algebra is being wedged, i.e. are there multiple wedges? Or is there just one wedge operator which should be wedging everything possible?



Is there a wedge which operates on multiple vector spaces?
$$\Lambda(V)\otimes\Lambda(W)\cong \Lambda(V\oplus W)$$
where the the wedge on the right hand side corresponds to your
$$\wedge\_{all} : (\Lambda^p(V)\otimes\Lambda^q(W))\times (\Lambda^r(V)\otimes\Lambda^s(W))\to \Lambda^{p+r}(V)\otimes\Lambda^{q+s}(W)$$
on the left hand side
with the correct sign conventions. 


Also your first operation is well defined, but you need a use for it.


