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?