## tensor products – Natural isomorphisms between vector spaces

Let $$X, Y$$ and $$Z$$ be any three finite dimensional vector spaces s.t. $$X$$ and $$Y$$ have the same dimension. Now, let assume there exists a natural isomorphism between $$Hom(X, Z)$$ and $$Hom(Y, Z)$$ (i.e., $$Hom(X, Z) cong Hom(Y, Z)$$). What can be conclude on $$X$$ and $$Y$$?
Does it imply that $$X cong Y$$, i.e., there exists a natural isomorphism between them?

I encounter this in a proof showing that $$(U otimes V) otimes W cong U otimes (V otimes W)$$.
I can derive all the details showing that $$Hom ((U otimes V) otimes W, Z) cong Hom( U otimes (V otimes W), Z)$$. Unfortunately, I cannot derive the final step.

## functional analysis – Extension of essentially self-adjoint operators on tensor products

The content of my question is based on Reed & Simon. Let $$A$$ be a given densely-defined self-adjoint operator on a Hilbert space $$mathscr{H}$$. Let $$mathcal{D}_{A}$$ be a domain of self-adjointness of $$A$$. Thus, the restriction of $$A$$ to $$mathcal{D}_{A}$$ is essentially self-adjoint, which means that its closure $$bar{A}$$ is self-adjoint. Here, we could take $$mathcal{D}_{A}$$ to be the domain $$D(A)$$ of $$A$$ itself, in which case $$bar{A} = A$$, since $$A=A^{*}$$ is closed.

Now, let $$A_{1},…,A_{n}$$ be (densely-defined) self-adjoint operators on $$mathscr{H}_{1},…mathscr{H}_{n}$$ respectivelly. For each $$A_{i}$$, let $$mathcal{D}_{A_{i}}$$ be a domain of self-adjointness of $$A_{i}$$ and set $$mathcal{D}_{A_{1}}otimes cdots otimes mathcal{D}_{A_{n}}$$ to be the set of all finite linear combinations of elements of the form $$varphi_{1}otimes cdots otimes varphi_{n}$$, with $$varphi_{i}in D_{A_{i}}$$. In addition, let $$A_{pi} := A_{1}otimes cdots otimes A_{n}$$ be defined on $$mathcal{D}_{A_{1}}otimes cdots otimes mathcal{D}_{A_{n}}$$ as follows. For each $$varphi_{1}otimes cdots otimes varphi_{n}$$, set:
$$(A_{1}otimes cdots otimes A_{n})(varphi_{1}otimes cdots otimes varphi_{n}) := A_{1}varphi_{1}otimes cdots otimes A_{n}varphi_{n}$$
and extend it to $$mathcal{D}_{A_{1}}otimes cdots otimes mathcal{D}_{A_{n}}$$ by linearity. Finally, if $$I_{i}$$ denotes the identity operator on $$mathscr{H}_{i}$$, let $$A_{Sigma}$$ be defined on $$mathcal{D}(A_{1})otimes cdots otimes mathcal{D}(A_{n})$$ by:
$$A_{Sigma}:= A_{1}otimes I_{2}otimes cdots otimes I_{n} + I_{1}otimes A_{2}otimes cdots otimes I_{n}+ cdots + I_{1}otimes I_{2}otimes cdots otimes A_{n}$$
As proved in Reed & Simon, $$A_{Sigma}$$ is essentially self-adjoint on its domain. This implies that $$A_{Sigma}$$ has a self-adjoint extension. But, as before, we could take $$D_{A_{i}}= D(A_{i})$$ in which case I’d expect that the self-adjoint extension of $$A_{Sigma}$$ would be itself. Is this correct? And, if so, is there a easy way to prove it?

## replacement – commutator between tensor products

I am interested in some quantum mechanical calculations in Mathematica. I am working with a tensor product of Hilbert spaces $$mathcal H = mathcal H_1otimes mathcal H_2$$. I would like to implement the following operator rule, to be applied in any occurrence in my calculations:
$$[hat A_1,hat B_1otimes hat A_2] = [hat A_1,hat B_1]otimes hat A_2$$
where the indices refer to the two Hilber spaces. Which is the most convenient way to do so?

## Rlim versus tensor product

Let $$R$$ be a coherent ring, and let $$(M_n)_{ngeq 1}$$ and $$(N_n)_{ngeq 1}$$ be two inverse systems of finitely generated flat $$R$$-modules. If $$R^1 lim M_n=R^1 lim N_n = 0$$, is it true also that $$R^1 lim (M_n otimes_R N_n)=0$$? This seems very reasonable but I’m having trouble seeing it.

A more conceptual question: is there a reasonably general "Kunneth formula" for $$Rlim$$ on inverse systems of $$R$$-modules?

## commutative algebra – Galois action on a tensor product of fields

Let $$K$$ be a field, $$F$$ a finite field extension of $$K$$ and let $$L$$ be an algebraic closure of $$K$$. Let $$G_K:=operatorname{Gal}(K^{text{sep}}|K)$$ be the absolute Galois group of $$K$$ whose action extends to $$L$$. We have
$$Fotimes_K L=prod_{nu}{L}$$
where the product is indexed over field embeddings $$nu:Fto L$$. Any $$sigmain G_K$$ acts on the left-hand side of this equality by $$1otimes sigma$$. What is the corresponding action on the right-hand side in terms of Galois groups?

Many thanks!

## algebraic geometry – Is the tensor by an invertible sheaf functor fully faithful?

My question arises while trying to prove that given a coherent $$mathcal{O}_X$$-module $$F$$ and an invertible locally free sheaf of rank $$1$$ (line bundle) $$L$$, then $$H^0(X,Fotimes L^n)cong Hom(L^{-n},F).$$

Here $$L^n$$ denotes the $$n$$-th tensor power of $$L$$, and the minus sign indicates its inverse in the Picard group of $$X$$. I’d like to prove that using the tensor-Hom adjunction and other two facts that maybe very simple but I’m not so sure about. My attempt would be so to consider isomorphisms
$$Hom(-,Hom(L^{-n},F))cong Hom(L^{-n}otimes -,F)cong Hom(-,L^notimes F)cong Hom(-,H^0(X,L^notimes F)),$$
and eventually use Yoneda embedding. Are the two last step correct? That is $$(i)$$ tensor product functor $$Lotimes -$$ is fully faithful and $$(ii)$$ there is isomorphism between groups $$Hom_{mathcal{O_{X}}}(-,G)$$ and $$Hom_{Ab}(-,G(X))$$ for a sheaf of modules $$G$$? If so, how to prove them? If not, how to prove the original statement?

## rt.representation theory – Infinite dimensional irreducible representations of a tensor product

Let $$A = langle frac{mathrm{d}}{mathrm{d} X}, X rangle$$ be the Weyl algebra over a field $$k$$ and let $$B = k(X)$$. Take $$M = k(X)$$, regarded as a left $$A otimes_k B$$-module by linear extension of

$$bigl(theta otimes g(X) bigr) f(X) = theta bigl( g(X)f(X) bigr)$$

for $$theta in A$$ and $$g(X) in k(X)$$. This is bilinear in $$theta$$ and $$g(X)$$ and $$(1_A otimes 1)f(X) = f(X)$$, so $$M$$ is a well-defined $$A otimes k(X)$$-module. The restriction of the action to the subalgebra $$A otimes_k k cong A$$ makes $$M$$ the natural representation of the Weyl algebra $$A$$. This is known to be irreducible. Therefore $$M$$ is irreducible.

The restriction of $$M$$ to the subalgebra $$k otimes_k k(X) cong k(X)$$ is the free $$k(X)$$-module. So if $$M cong V otimes_k W$$ with $$A$$ acting on $$V$$ and $$k(X)$$ acting on $$W$$ then $$W$$ is free. Hence $$W$$ is not irreducible. Therefore there is no factorization of $$M$$ as $$V otimes_k W$$ where $$V$$ is an irreducible (necessarily infinite-dimensional) representation of $$A$$ and $$W$$ is an irreducible representation of $$k(X)$$.

## dg.differential geometry – About an explicit formula of the curvature tensor by holomorphic sectional curvatures

Let $$(M, g)$$ be a Riemannian manifold. Define the curvature tensor convention as follows.

$$R(X, Y) Z = nabla_X nabla_Y Z – nabla_Y nabla_X Z – nabla_{(X,Y)} Z$$

$$R(X,Y,Z,W) = g(R(X,Y)Z, W)$$

It is well-known that the curvature tensor $$R$$ is explicitly expressed by the sectional curvatures. This can be found in Jost’s
Riemannian Geometry and Geometric Analysis.

Lemma. With $$K(X,Y) = R(X, Y, Y, X)$$,
begin{align} R(X,Y,Z,W) = & + K(X+W, Y+Z) – K(X+W, Y) – K(X+W, Z) \ &- K(X, Y+Z) – K(W,Y+Z) + K(X,Z) + K(W,Y) \ & – K(Y+W, X+Z) + K(Y+W, X) + K(Y+W,Z) \ & + K(Y,X+Z) + K(W,X+Z) – K(Y,Z) – K(W,X).end{align}

My question is whether there is a similar explicit formula for the Riemannian curvature tensor and the holomorphic sectional curvature when we assume that the manifold is KÃ¤hler.

I think this MO question is related to my question. From this polarization formula, at least, we know that the curvature tensor is determined algebraically by holomorphic sectional curvatures. However, I am curious about whether there is a (relatively) simpler formula describing the algebraic relation between the curvature tensor and holomorphic sectional curvatures.

Thanks!

## ct.category theory – tensor triangulated categories

Let $$(mathcal{T},otimes)$$ be a monoidal (not necessarily symmetric and possibly without unit object) triangulated category where $$-otimes-$$ is exact on both variables. Let $$S$$ be a set of objects in $$mathcal{T}$$ such that for any $$s_1in S$$ and $$s_2in S$$ we have that $$s_1otimes s_{2}in S$$.

I was wondering if the thick subcategory generated by $$S$$ is automatically monoidal?

## galois theory – How to express a tensor product of algebras as a product of fields

I am dealing with this question in one of my courses:

Which of the following algebras are fields? Products of fields? Describe these fields.

1. $$mathbb{Q}(sqrt(3){2}) space otimes_{mathbb{Q}} space mathbb{Q}(sqrt{2})$$
2. $$mathbb{Q}(sqrt(4){2}) space otimes_{mathbb{Q}} space mathbb{Q}(sqrt{2})$$
3. $$mathbb{F}_2(sqrt{T}) space otimes_{mathbb{F}_2(T)} mathbb{F}_2(sqrt{T})$$
4. $$mathbb{F}_4(sqrt(3){T}) space otimes_{mathbb{F}_4(T)} mathbb{F}_4(sqrt(3){T})$$

I know that any finite algebra have finitely many maximal ideals.

Say $$m_1,…,m_k$$ be the maximal ideals of our finite algebra A.
Then $$A cong frac{A}{m_1^{n_1}}times …times frac{A}{m_r^{n_r}}$$ for some $$n_iinmathbb{N}$$.

Hence if $$space forall i space ;n_i=1$$ , then A is a product of fields.

Also there are some helpful theorems that I’ve referred , in my answers document, which is linked bellow in my answer to this problem.