bio | website | |
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age | ||
visits | member for | 3 years, 1 month |
seen | Sep 2 '14 at 23:38 | |
stats | profile views | 207 |
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... |