2
$\begingroup$

This might be inappropriate for the MO-level. If so I'll delete it...

Suppose $V$ is a $\mathbb{Z}$-graded vector space and $\overline{T}(V):=V \oplus V\otimes V \oplus \otimes^3 V \ldots$ is the 'reduced tensor power' of $V$ in the graded sense.i.e:

$V\otimes V:=\oplus_{z\in\mathbb{Z}}\oplus_{p+q=z}V_p\otimes V_q$

We don't see $T(V)$ as a (co)algebra, just as a plain graded vector space.

Then on a first "empirical observation" I would say, that if we apply the reduced tensor functor again, i.e make $\overline{T}(\overline{T}(V))$ then this is naturally isomorphic to $\overline{T}(V)$ because "all tensor powers are already there".

So this should be a consequence of the 'freeness' of $\overline{T}(V)$, but I can't see how this can be proofen.

$\endgroup$
4
  • 2
    $\begingroup$ Assume that each homogeneous component of $V$ is free of finite rank over $\mathbb Z$. Assume also that the $0$-th homogeneous component of $V$ is $0$. Then, if $H\left(W\right)$ denotes the Hilbert series of a graded $\mathbb Z$-module $W$ whose homogeneous components are free, then we have $H\left(\overline T\left(V\right)\right) = \frac{H\left(V\right)}{1-H\left(V\right)}$. From this, it should be pretty easy to see that in general, $H\left(\overline T\left(V\right)\right) \neq H\left(\overline T\left(\overline T\left(V\right)\right)\right)$. $\endgroup$ Jul 22, 2013 at 15:51
  • $\begingroup$ I'm not sure I understand this.You mean if the underlying field of the vector space is finite? In that case ok, I have to rewrite the question, because I'm only interested in $\mathbb{R}$-vector spaces. $\endgroup$
    – Nevermind
    Jul 22, 2013 at 16:11
  • $\begingroup$ To make darij's comment more explicit: Let $V_1 = \mathbb{R}$ and all other $V_0=0$. Then the graded pieces of $\bar{T}(V)$ have dimensions $(0,1,1,1,1,\dots)$ and the graded pieces of $\bar{T}(\bar{T}(V))$ have dimensions $(0,1,2,4,8,16, \dots)$. This simply isn't true. $\endgroup$ Jul 22, 2013 at 17:39
  • $\begingroup$ ok. I see now . $\endgroup$
    – Nevermind
    Jul 22, 2013 at 18:15

1 Answer 1

3
$\begingroup$

If $V$ is $\mathbb Z$-graded, then $\overline{T}(V)$ is $\mathbb Z_{\ge1}\times \mathbb Z$-graded with $\overline{T}(V)_{p,q} = \bigoplus_{z_1+z_2+\dots+z_p = q}V_{z_1}\otimes V_{z_2}\otimes\dots\otimes V_{z_p}$.

$\overline{T}$ is a functor from $\mathbb Z$-graded vector spaces to $\mathbb Z_{\ge1}\times \mathbb Z$-graded algebras which is left adjoint to the forgetful functor, where one has to pay attention to the degrees ...

$\endgroup$
3
  • $\begingroup$ And by the way: $\overline{T}$ is in addition a functor into the category of locally nilpotent graded coassociative coalgebras and there it is right adoint to the forgetful functor $\endgroup$
    – Nevermind
    Jul 22, 2013 at 16:06
  • $\begingroup$ But as I said, the graded tensor power makes $\overline{T}(V)$ into a $\mathbb{Z}$-graded vector space by $\overline{T}(V)_j:=\oplus_{k}\oplus_{p_1,\ldots,p_k=j}V_{p_1}\otimes \cdots\otimes V_{p_k}$, where each $p_h\geq 1$. $\endgroup$
    – Nevermind
    Jul 22, 2013 at 16:16
  • $\begingroup$ You decided to ignore the grade $k$ which exactly encodes your feeling that "all the tensor powers are already there". But in $\overline{T}\overline{T}(V)$ they are there multiple times; you then have 3 degrees and cannot reduce them to 2 correctly. $\endgroup$ Jul 24, 2013 at 5:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.