changes to hw 1

By / / Uncategorized / 11 comments

Hi all,

I am postponing problems 4(c) and 4(d) until HW2, since we haven’t defined the necessary concepts yet.

In problem 3, all you need to know about {\mathbb F}G is that we defined \Delta:{\mathbb F}G \rightarrow {\mathbb F}G \otimes {\mathbb F}G by  setting \Delta(g) = g \otimes g for g \in G, and then extending linearly.

In problem 5, just as T(V) is decomposed as a direct sum of V^{\otimes k} (tensors of order k) for each k, there is a similar decomposition for S(V) and \wedge(V). The Hilbert series for these other two algebras is defined analogously. (You can start with a basis x_1, \ldots, x_d, find a basis of each of these three spaces. In each case, count how many basis vectors are order k tensors for each k, and then find the resulting generating function.)

11 comments

  1. Lisa Clayton
    Reply

    In problem 5a, should \sum_{n=0}^{\infty} be \sum_{k=0}^{\infty}?

  2. Duquec
    Reply

    Question: Are we allowed to assume that vector spaces (that are not algebras) have finite dimension? So, for example, are the \mathbb{F}-vector spaces in Problem 1 finite dimensional?

    • federico
      Reply

      no, and my feeling is that if you have a proof in the finite-dimensional case, it won’t be difficult to extend it to the general case.
      (keep in mind that many fundamental algebras (and even many combinatorial ones) are infinite-dimensional, such as the tensor algebra, the symmetric algebra, the exterior algebra, the algebra of shuffles…)

  3. spineda6239
    Reply

    For problems b) and c) are the relations we’re modding out by only involving {V^{\otimes 2}}?

    For example, is the relation in c) saying {v_{i}} \otimes {v_{i}} \otimes ... \otimes {v_{i}} \otimes... in general or only with two components {v_{i}} \otimes {v_{i}}?

    For b), is the relation saying

    ( {v_{1}} \otimes {v_{2}} \otimes ... \otimes {v_{j}} \otimes {v_{i}} \otimes ... ) - ( {v_{1}} \otimes {v_{2}} \otimes ... \otimes {v_{j}} \otimes {v_{i}} \otimes ... )

    I feel like the Hilbert series for b) and c) make more sense if the relations includes all {V^{\otimes k}}, right?

  4. katrina888
    Reply

    I thought it was for any v_i \otimes v_j…So like for 2c, in S^3(V), v_1 \otimes v_1 \otimes v_2 = 0, not just v_1 \otimes v_1 \otimes v_1?

  5. spineda6239
    Reply

    Hey Katrina,

    Did you mean 5c? Also, the relationship stated says {v} \otimes {v}, so I think it means the same {v} component.

  6. Duquec
    Reply

    Since v\otimes v = 0, then I think that it follows that v\times v \otimes v = (v\otimes v) \otimes v = 0\otimes v = 0.

  7. federico
    Reply

    In problem 5, keep in mind that we are modding out by ideals (not just subspace). So, for example, every multiple of v \otimes v is in the ideal we mod out by to define the exterior algebra \wedge(V). You can think of this as Katrina and Steven (duquec) did: in \wedge(V), always feel free to plug in v \otimes v = 0. Similarly, in the symmetric algebra, always feel free to plug in u \otimes v = v \otimes u.

    While I am at it, let me warn you of an easy thing to miss about the exterior algebra: if you choose a basis \{v_i\} of V, then you are not only setting v_i \otimes v_i = 0 for the elements of the basis, but also for any other v \in V. For instance, (v_i+v_j) \otimes (v_i+v_j) = 0 for all i,j; what does that imply?

  8. Duquec
    Reply

    Hmm . . . I missed that while counting the dimension of the algebra V^{\otimes k} / \langle v\otimes v : v\in V \rangle to get a closed form for the generating function.

    (v_i + v_j)\otimes(v_i + v_j) = 0 implies that v_i\otimes v_j = -v_j\otimes v_i because

    (v_i + v_j)\otimes(v_i + v_j) = v_i \otimes v_i + v_i \otimes v_j + v_j \otimes v_i + v_j \otimes v_j = 0

    v_i \otimes v_j + v_j \otimes v_i = 0

    v_i \otimes v_j = -v_j \otimes v_i.

    I remember, in my past life as a physics major, that the vector cross product had something to do with exterior algebras. I also remember that the term “anti-commutative” came up along the way. I have the feeling that the cross product is an example of an exterior algebra of \mathbb{R}^3. This would be nice because it would be the first example, to my knowledge, of an algebra where a tensor symbol “does something” –in other words, putting an “\otimes” between two things actually yields something other than an “\otimes” between two things.

    (Finally, another interesting “fact” (?) about the cross product is that it only exists in three dimensions and seven dimensions, whatever that means.)

  9. Duquec
    Reply

    Two new things: (1) I take back my comment saying that cross product is an instance where the “\otimes” symbol does something. This is actually inaccurate because I feel that we can think of this symbol as doing something in several of the examples that have been covered in class.

    (2) As a complement to my somewhat shallow observation concerning what (v_i + v_j)\otimes (v_i + v_j) is: Notice how similar this relation is to the one we are modding out by in part (b) –in this case we have sum instead of a subtraction. In any case, we should “get rid of” these sums in the tensor algebra T(V)

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