# Algebra Problems

1. Feb 6, 2009

### 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 have$e_{ij}^* \wedge e_j^* = e_i^* \wedge (e_j^* \wedge e_j^*)=0$ 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

2. Feb 7, 2009

### CompuChip

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
$$f(a v + w, u) = (a v + w) \wedge u = a (v \wedge u) + w \wedge u$$

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.

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 $n > k$.

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.

3. Feb 7, 2009

### 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?

4. Feb 7, 2009

### slider142

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.

The nth exterior power of a vector space is the vector space of all n-vectors, rigorously, elements of Vk = $V\times\dots\times V$, 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.

Last edited: Feb 7, 2009
5. Feb 8, 2009

### latentcorpse

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