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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.

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    $\begingroup$ 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, $H\left(\overline T\left(V\right)\right) \neq H\left(\overline T\left(\overline T\left(V\right)\right)\right)$. $\endgroup$
    – darij grinberg
    Commented 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
    Commented 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. $\endgroup$
    – David E Speyer
    Commented Jul 22, 2013 at 17:39
  • $\begingroup$ ok. I see now . $\endgroup$
    – Nevermind
    Commented Jul 22, 2013 at 18:15

1 Answer 1

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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 ...

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  • $\begingroup$ 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
    Commented 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
    Commented 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. $\endgroup$
    – Peter Michor
    Commented Jul 24, 2013 at 5:17

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