changes to hw 1

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.)

1. Lisa Clayton

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

• yes, thanks!

2. Duquec

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?

• 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. 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

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. 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

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. 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

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

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 $T(V)$