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bio website math.harvard.edu/~oantolin
location Cambridge, Massachusetts, USA
age 34
visits member for 4 years, 2 months
seen 18 hours ago

1d
reviewed Approve suggested edit on Prove that a convex $d$-polytope has at least $d+1$ facets
1d
reviewed Approve suggested edit on Coefficients in Representer theorem
1d
reviewed Approve suggested edit on Tricky limit of summation
1d
reviewed Approve suggested edit on Explicit probability for vertical 2D percolation
2d
comment Morphisms in the category of natural transformations?
It only makes sense to talk about representing functors $G \to \mathrm{Set}$, the codomain of a representable functor is $\mathrm{Set}$, @NikolajK.
2d
comment Morphisms in the category of natural transformations?
I don't understand your comment, @NikolajK, in particular I don't know where you got the $\hom(G,G)$ from. In the last two lines I say that for $f,g : G \to H$ we have $\mathrm{Nat}(f,g) = \{ y \in H : \forall x\in G, f(x) = y g(x) y^{-1}\}$ which is very easy to see: each natural transformation has a single component $y$ (because $G$ has a single object) and the stated condition is just naturality.
2d
comment Loop of a topological group acting on different points being homotopic to constant maps.
You don't need path connectedness of $G$ since the only portion of $G$ you ever use is the image of $\alpha$, which is contained in the path component of the identity of $G$.
2d
revised Loop of a topological group acting on different points being homotopic to constant maps.
added 526 characters in body
2d
answered Loop of a topological group acting on different points being homotopic to constant maps.
Oct
24
comment Prove that theorem about trace of non-negative matrix:
With that definition of "nonnegative matrix", @Dr.Mundo, the result is obviously false (take a matrix with 0 on the diagonal and 1 off the diagonal).
Oct
23
reviewed Approve suggested edit on the 0-th homology of a simplical complex
Oct
23
reviewed Approve suggested edit on Permutation Groups question in abstract algebra
Oct
22
awarded  category-theory
Oct
21
awarded  Explainer
Oct
21
comment Different Definitions of Tensor product, Halmos, Formal Sums, Universal Property
I did mean $V \otimes V \to \mathbb{R}$, that is, bilinear maps $V \oplus V \to \mathbb{R}$. Also, $(V \otimes W)^\ast \cong V^\ast \otimes W^\ast$ is true for infnite dimensional spaces, but for cardinal arithmetic reasons, not in any natural way. The map $V^\ast \otimes W^\ast \to (V \otimes W)^\ast$ given by $\alpha \otimes \beta \mapsto (v \otimes w \mapsto \alpha(v) \beta(w))$ is an isomorphisms when $V$ and $W$ are finite dimensional but not in general.
Oct
21
comment Different Definitions of Tensor product, Halmos, Formal Sums, Universal Property
I already mentioned one way to say what the variance is, but what you say is also related: given $x,y \in V$, you get an $x \otimes y$ of $V \otimes V$ (as you said, if you defined $V \otimes V$ as $\mathrm{Bil}(V,V;\mathbb{R})$, this element would be $w \mapsto w(x,y)$); but you don't canonically get an element of $(V \otimes V)^\ast$. If you pick an isomorphism $V \cong V^\ast$, you get a corresponding isomorphism $(V \otimes V)^\ast \cong (V^\ast \otimes V^\ast)^\ast$, and you can get an element of the latter space from $x$ and $y$: $\alpha \otimes \beta \mapsto \alpha(x) \beta(y)$.
Oct
21
comment Different Definitions of Tensor product, Halmos, Formal Sums, Universal Property
I think something different is the issue. In differential geoemtry people makes use of both maps $V \otimes V \to \mathbb{R}$ and elements of $V \otimes V$. The first are called 2-tensors (there are also $k$-tensors for all $k$) and the second are called bivectors (and for larger $k$, polyvectors). These tow spaces, $(V \otimes V)^\ast \cong V^\ast \otimes V^\ast$ (for finite dimensional $V$, which is the case that arises for finite dimensional manifolds), and $V \otimes V$ depend on $V$ with different variance. (to be cont'd)
Oct
21
comment Different Definitions of Tensor product, Halmos, Formal Sums, Universal Property
Taking just bilinear forms gives a functor of the wrong variance, @Stefan: given linear maps $U \to U'$ and $V \to V'$, what you naturally get is a linear map $\mathrm{Bil}(U',V';\mathbb{F}) \to \mathrm{Bil}(U,V; \mathbb{F})$. This is contravariant, but you want the tensor product to be a covariant functor.
Oct
21
answered Proof that the tensor product is the coproduct in the category of R-algebras
Oct
21
answered Different Definitions of Tensor product, Halmos, Formal Sums, Universal Property