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Suppose $V$ is a $\mathbb{Z}$-graded vector space and $T(V):=\oplus_{j\in \mathbb{Z}}\oplus_{p+q=j}V_p\otimes V_q$ the graded tensor power (as a vector space).

1) Is then $T(T(V))\simeq T(V)$?

2) What about the various quotients?

Suppose $\odot V$ is the 'graded symmetric' quotient. Is then $\odot(\odot V)\simeq \odot V$, too? And similarly for the graded anti-symmetric quotient $\wedge V$.

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  • $\begingroup$ Can you explain what a graded vector space is? I am only familiar with graded ring/module/algebra. In case of vector space the grading seems to carry no extra information. $\endgroup$ – Marek Jul 15 '13 at 12:59
  • $\begingroup$ See for example ncatlab.org/nlab/show/graded+vector+space. $\endgroup$ – Mark Neuhaus Jul 15 '13 at 13:11
  • $\begingroup$ In that case I am not sure I understand the question. When only finitely many $V_i$ are nontrivial $T(T(V))$ will have strictly higher dimension than $T(V)$, so they can't be isomorphic. And similarly in the symmetric and antisymmetric cases. $\endgroup$ – Marek Jul 15 '13 at 13:37
  • $\begingroup$ T(V) has infinite dimension and so as $T(T(V))$. Tis is already true in the non graded situation. $\endgroup$ – Mark Neuhaus Jul 15 '13 at 14:02
  • $\begingroup$ I don't see how that's so. If $V$ is trivial, then obviously $T(V)$ will be as well. Are you implicitly assuming that $V_p$ is non-trivial for every $p$ or something? $\endgroup$ – Marek Jul 15 '13 at 16:04
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No, there is no reason to expect any isomorphism like this (at least not a natural isomorphism; for some particular $V$ there will be unnatural isomorphisms that are basically coincidental). For instance, if $V_0=k$ (where $k$ is the field) and $V_n=0$ for $n\neq 0$, $T(V)$ is just the polynomial ring $k[x]$ concentrated in degree $0$ and then $T(T(V))$ is a noncommutative polynomial ring $k\langle x_0,x_1,\dots\rangle$ in infinitely many generators (with the generator $x_n$ corresponding to the basis element $x^n\in k[x]$). Similarly, in this case $\odot(V)$ is also $k[x]$ and $\odot(\odot(V))$ is a commutative polynomial ring $k[x_0,x_1,\dots]$ in infinitely many variables.

You might think that such an isomorphism should hold because $T(V)$ is "$V$ turned into an algebra", so $T(T(V))$ would be the same as $T(V)$ since $T(V)$ is already an algebra. The problem with this idea is that the definition of $T(T(V))$ does not know that $T(V)$ is an algebra; it only uses the vector space structure of $T(V)$. That is, $T(T(V))$ is the underlying vector space of $T(V)$ turned into an algebra as freely as possible. Since you are doing this as freely as possible, there is no reason that the multiplication should agree with the multiplication you already have on $T(V)$; instead, if you take two elements of $T(V)$ and multiply them in $T(T(V))$, you should expect the product to be some new purely "formal" product that is not an element of $T(V)$.

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