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seen Sep 2 at 23:38

Jul
2
awarded  Curious
Jul
22
comment Iterated Reduced Tensor Power of Graded Vector spaces
ok. I see now .
Jul
22
comment Iterated Reduced Tensor Power of Graded Vector spaces
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 $p_h\geq 1$.
Jul
22
comment Iterated Reduced Tensor Power of Graded Vector spaces
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.
Jul
22
comment Iterated Reduced Tensor Power of Graded Vector spaces
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
Jul
22
asked Iterated Reduced Tensor Power of Graded Vector spaces
Jun
2
comment how to make the category of chain complexes into an $(\infty,1)$-category
I didn't read Higher algebra because I thought it only presents the (oo,1)-category as a quasi-category not as a simplicial category.
Jun
2
comment how to make the category of chain complexes into an $(\infty,1)$-category
Yes I tink I'm more or less after 2.) ... But most likely this has already been done somewhere. And I would prefere to read that instead of calculating it by myself.
Jun
1
asked how to make the category of chain complexes into an $(\infty,1)$-category
May
25
comment Natural Isomorphism of $S(V[1])$ and $(\bigwedge V)[n]$
Seems like you are right. Unfortunately I don't speak French. Anyway the answer is ok.
May
25
accepted Natural Isomorphism of $S(V[1])$ and $(\bigwedge V)[n]$
May
24
awarded  Citizen Patrol
May
23
asked Natural Isomorphism of $S(V[1])$ and $(\bigwedge V)[n]$
May
20
accepted Vector field pull back from embedding
May
20
comment Vector field pull back from embedding
Ok. Good point. So this "pullback" depends on the choice of $r$.
May
20
revised Vector field pull back from embedding
added 341 characters in body; added 1 characters in body
May
20
comment Decalage isomorphism and algebra structure
Did you read the paper arxiv.org/abs/math/0601312 ?
May
19
comment Vector field pull back from embedding
Don't need a preferred base-point I think. To be functorial, it should be enough to show that any choice of the retract gives the same vector field on $M$, but as I said at any $f(m)$ the equation $r\circ f=id_M$ gives $r'_*(x)=r_*(x)$ for any two $r'_*,r_*:T_{f(m)}N \to T_mM$.
May
19
comment Vector field pull back from embedding
If $dim(M)=dim(N)$ then $r=f^{-1}$.
May
19
comment Vector field pull back from embedding
Since $r\circ f=id_M$ every choice of $r$ should give the same $r_*$ on $f(m)$. And in the equi-dimensional case this is just the ordinary pullback of a vector field along a diffeomorphism, since embeddings are diffeomorphisms then...