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Suppose $V$ is a vector space and $T(V)$ the tensor algebra of $V$. What happens if we take $T(T(V))$ that is the tensor algebra of the (vector space) $T(V)$?

I 'guess' I heard that $T(T(V)) \simeq T(V)$ but I can't find it anywhere, so maybe this was just a dream....

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1 Answer 1

You should realise that the application of the outer $T$ in $T(T(V))$ implicitly demotes $T(V)$ from an algebra to a vector space. Then one does not have $T(T(V)) \simeq T(V)$ as an algebra in general (they are isomorphic as vector spaces though; for instance if $V$ is finite dimensional, they are both countably-infinite dimensional vector spaces).

For instance, if $V$ is finite dimensional, then the algebra $T(V)$ has the property of possessing a finite dimensional subspace that generates it as an algebra, viz. the copy of $V$ inside $T(V)$. However $T(T(V))$ does not have this property, because the elements of $T(V)$ inside $T(T(V))$ form an infinite dimensional subspace of irreducible elements (the only decompositions for them involve a nonzero scalar of the field; this is a consequence of the "demotion" referred to above).

As a graded algebra this is even more evident: $T(V)$ has finite-dimensional homogeneous components in every degree, while $T(T(V))$ is infinite-demensional in any positive degree.

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As far as know the category of algebras IS monoidal, so taking the tensor algebra of an algebra makes sense. –  Mark Neuhaus May 17 '12 at 11:33
    
Since $T(V)$ is a free algebra it boils down to the question whether or not the free algebra B<A<X>> of a free algebra A<X> over a set X is equivalent to A<X> –  Mark Neuhaus May 17 '12 at 11:49
    
Or to be less abstract if we have two elements of the tensor algebra $T(V)$,lets say $x_1 \otimes \ldots \otimes x_k$ and $y_1 \otimes \ldots \otimes y_l$ then their tensor product in $T(T(V))$ is $x_1 \otimes \ldots \otimes x_k \otimes y_1 \otimes \ldots \otimes y_l$ and this is again an element of $T(V)$. –  Mark Neuhaus May 17 '12 at 11:57
    
$T(T(V))=T(W)$, where $W=T(V)$, makes sense. The point here is that the identification $T(W)\simeq T(V)$ does not hold at the algebra level. –  Avitus Jul 15 '13 at 12:25
    
@user33433: I think you are confusing the tensor algebra and the symmetric algebra. The tensor algebra is non commutative. Taking an algebra over a non commutative ring makes no sense, since one wants multiplication in an $R$-algebra to be $R$-bilinear (without that it would be rather pointless to talk about an $R$-algebra), but that is not possible unless $R$ is commutative. –  Marc van Leeuwen Oct 7 '13 at 9:50

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