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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?

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While Peter has probably answered your question, I'd like to stress that when defining an operator it is imperative to specify what exactly it does, not only where it takes its values and where it has its domain. This lack of detail, along with some ambiguous wording (how are you "wedging" an algebra here and what does it even mean?) makes this question harder to understand than it should be. –  darij grinberg Mar 27 '13 at 21:54
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Either way, you can definitely construct both a bilinear map $ \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)$ which sends $\left(\left(v_1\wedge v_2\wedge ...\wedge v_p\right)\otimes\left(w_1\wedge w_2\wedge ...\wedge w_q\right), \left(x_1\wedge x_2\wedge ...\wedge x_r\right)\otimes \left(y_1\wedge y_2\wedge ...\wedge y_s\right)\right)$ to ... –  darij grinberg Mar 27 '13 at 21:56
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... $\left(v_1\wedge v_2\wedge ...\wedge v_p\wedge x_1\wedge x_2\wedge ...\wedge x_r\right) \otimes \left(w_1\wedge w_2\wedge ...\wedge w_q\right)\otimes \left(y_1\wedge y_2\wedge ...\wedge y_s\right)$, and a bilinear map $\wedge_{all} : (\Lambda^p(V)\otimes\Lambda^q(W))\times (\Lambda^r(V)\otimes\Lambda^s(W))\to \Lambda^{p+r}(V)\otimes\Lambda^{q+s}(W)$ which sends the same thing to ... –  darij grinberg Mar 27 '13 at 21:57
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... $\left(v_1\wedge v_2\wedge ...\wedge v_p\wedge x_1\wedge x_2\wedge ...\wedge x_r\right) \otimes \left(w_1\wedge w_2\wedge ...\wedge w_q\wedge y_1\wedge y_2\wedge ...\wedge y_s\right)$. I am pretty sure that both of these maps are useful, so yes, you should specify which one you're meaning (if this is your question). –  darij grinberg Mar 27 '13 at 21:58
    
See also mathoverflow.net/questions/17335/… –  Douglas Zare Mar 28 '13 at 4:55
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up vote 5 down vote accepted

$$\Lambda(V)\otimes\Lambda(W)\cong \Lambda(V\oplus W)$$ 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.

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