Tensor Algebra of Tensor Algebra 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....
 A: 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.
