Solving Algebra Problems: Bilinear Multiplication & Supercommutativity

[latentcorpse]

ok. today we discussed the following:

The wedge product defines a bilinear multiplication in \Lambda(V^{*}) which is associateive i.e. (\alpha \wedge \beta) \wedge \gamma = \alpha \wedge (\beta \wedge \gamma) \forall \alpha,\beta,\gamma \in \Lambda(V^{*}) and which is supercommutative i.e. if \alpha^{k} \in \Lambda^{k}(V^{*}),\beta^{l} \in \Lambda^{l}(V^{*}) then \alpha^{k} \wedge \beta^{l} = (-1)^{kl} \beta^{l} \wedge \alpha^{k}

Question 1: What is a bilinear multiplication?
Question 2: Is my definition of supercommutative ok? I wasn't sure if it was (-1)^{kl} or (-1)^{k+l}? What does supercommutative mean?

Then as an example of this, we considered the following:

\alpha=e_1^* \wedge e_2^* - 3 e_2^* \wedge e_4^* = e_{12}^* -3 e_{24}^* \in \Lambda^{2}(V^{*}), \beta=3 e_1^* + 4e_2^*

Then,

\alpha \wedge \beta=6 e_{12}^* \wedge e_1^* +8 e_{12}^* \wedge e_2^* - 9 e_{24}^* \wedge e_1^* - 12 e_{24}^* \wedge e_2^*

but we havee_{ij}^* \wedge e_j^* = e_i^* \wedge (e_j^* \wedge e_j^*)=0 so 3 terms drop out and we're left with

\alpha \wedge \beta =- 9 e_{24}^* \wedge e_1^* = -9 e_2^* \wedge e_4^* \wedge e_1^8=\mathbf{- (-1)^{2} e_1^* \wedge e_2^* \wedge e_4^*=-e_{124}^*}

Question 3: Is \beta \in \Lambda^{1}(V^{*})?
Question 4: I don't understand how he got the part in bold. I have a feeling he just forgot to write in the 9 but the rearranging of that step uses the supercommutative algebra described above and I don't follow the logic there?

Any help would be greatly appreciated

cheers

 

[CompuChip]

I don't really know much about this topic, but since nobody has replied yet I will try to help you with my general knowledge

Question 1: What is a bilinear multiplication?

If you consider for example a vector space V, then a bilinear function on V is a function f on V x V which is linear in both its arguments. So if v, w, u are elements of V and x is a number, then
f(a v + w, u) = a f(v, u) + f(w, u)
f(v, a w + u) = a f(v, w) + f(v, u)

You can see the wedge product as a function of two arguments:
f(v, w) = v \wedge w
then bilinearity simply means that for example
f(a v + w, u) = (a v + w) \wedge u = a (v \wedge u) + w \wedge u

Question 2: Is my definition of supercommutative ok? I wasn't sure if it was (-1)^{kl} or (-1)^{k+l}? What does supercommutative mean?

I think so. Supercommutative probably either refers to the fact that you are commuting a k- and an l-form, or to the fact that it isn't precisely a commutation but some generalization where an extra sign may occur. Since I recall something like
v \wedge w = - w \wedge v
when v and w are one-forms, I think it should be (-1)^(k l). That is: if both arguments are even or odd in grade you can switch them, if one is even and the other odd you get a minus sign.

Question 3: Is \beta \in \Lambda^{1}(V^{*})?

Yes. You can see it by the fact that it is expressed in the basis \{ e_1^*, e_2^* \}. For a two-form you would need a basis like \{ e_{ij} = e_i \wedge e_j \mid i < j \} (I am dropping the asterisks, too much work to put them in), for an n-form in general \{ e_{i_1 \cdots i_n} = e_{i_1} \wedge \cdots \wedge e_{i_n} \mid i_1 < \cdots < i_n \}. (Note that this implies that if V is k-dimensional, there are no n-forms for n > k.

Question 4: I don't understand how he got the part in bold. I have a feeling he just forgot to write in the 9 but the rearranging of that step uses the supercommutative algebra described above and I don't follow the logic there?

Yes, it looks like he just forgot the 9. He commutes the e1 to the front, to have all the elements in a logical order.

 

[latentcorpse]

thanks a lot. quick questions though.

(i)if you agree with me that \alpha \in \Lambda^2(V^*),\beta \in \Lambda^1(V^*), then why don't we write them as \alpha^2 and \beta^1 since at the start of the definition we were considering an \alpha^k \in \Lambda^k(V^*) etc?

(ii)If we are calling them \alpha^k just because it's in that set, how do we distinguish it from being \alpha \cdot ... \cdot \alpha k times?

(iii) What is \Lambda^{n}(V^*) - is it a set?

thanks for your help.

 

[slider142]

thanks a lot. quick questions though.

(i)if you agree with me that \alpha \in \Lambda^2(V^*),\beta \in \Lambda^1(V^*), then why don't we write them as \alpha^2 and \beta^1 since at the start of the definition we were considering an \alpha^k \in \Lambda^k(V^*) etc?

(ii)If we are calling them \alpha^k just because it's in that set, how do we distinguish it from being \alpha \cdot ... \cdot \alpha k times?

I have a tiny bit of experience here as well, from a very basic study of exterior calculus in diffy geo. The only reason we used \alpha^k in the previous scenario was to illustrate supercommutativity. It does not represent the wedge product of the same vector with itself k times. From the definition of the wedge product, \alpha^k = \alpha\wedge\dots\wedge\alpha for any k > 1 is the 0 vector.

(iii) What is \Lambda^{n}(V^*) - is it a set?

thanks for your help.

The nth exterior power of a vector space is the vector space of all n-vectors, rigorously, elements of V^k = V\times\dots\times V, quotiented by a subspace defined by the wedge product. That is to say, each vector in \Lambda^{2}V is an ordered pair of vectors in V, quotiented by the subspace made up of vectors of the form {(u+v, w) - [(u,w) + (v,w)], (u, v+w) - [(u,v) + (u,w)], a(u, v) - (au, v), (u, bv) - b(u, v), (u, u)}. As an example, if you regard R (the set of real numbers) as a vector space, you can regard the vector space \Lambda^{3}R as a certain quotient space of R³, albeit with a strange wedge product structure and having all vectors with any identical components identified with the 0 vector. However, this construction is more useful in the development of tensors and tensor calculus, where we do operations on many vectors per tensor.

 

[latentcorpse]

ok cheers. why is that vector space quotiented by that subspace though?