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?

Really thanks in advance 🙂

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 Kähler.

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.


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.