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-2
votes
0answers
112 views

Analogue of exponential exact sequence in mixed characteristic [closed]

Let $R$ be a discrete valuation ring with residue field of positive characteristic but fraction field of characteristic zero and $X_R$ a flat, projective $R$-scheme. Assume further that $X_R$ is ...
6
votes
2answers
389 views

What are the basic possibilities for a tensor product of two fields?

Let $k$ be a field, with $F,k'$ field extensions of $k$. The ring $k' \otimes_k F$ is denoted by $F_{k'}$. In Borel's Linear Algebraic Groups, it is claimed (I believe erroneously) that "each of ...
7
votes
1answer
226 views

Recovering a smooth manifold from its tensor fields

1) Consider the algebra $\mathcal T (M) = \bigoplus \limits _{p, q \ge 0} \mathcal T ^{p, q} (M)$, where $\mathcal T ^{p, q} (M)$ is the space of tensor fields of type $(p,q)$. I should endow it with ...
1
vote
0answers
39 views

Decomposing a matrix into the tensor product of a permutation and orthogonal matrix

Suppose I have a square matrix $A \in \mathbb{R}^{mn \times mn}$. I want to find $$\arg \min_{P, Q} \|A - P \otimes Q\|_F$$ where $P$ is an $m \times m$ permutation matrix and $Q$ is an $n \times ...
1
vote
0answers
31 views

Most accurate separation of tensor product of matrices

Given a matrix $A \in \mathbb{R}^{n^m \times n^m}$, how I can find a set of $m$ matrices $\{B_i\}$ such that $$\arg \min_{\{B_i\, \in\, \mathbb{R}^{n \times n}\}} \|A - B_1 \otimes \ldots \otimes ...
6
votes
0answers
188 views

Do there exist “non-algebraic tensor products” for “algebraic” triangulated categories?

Let us call a triangulated category algebraic if it admits a differential graded enhancement (i.e., an enrichment in complexes of abelian groups). Certainly, there is a notion of a tensor product on ...
2
votes
0answers
33 views

Tensor of Relative Bases

Suppose $U,V,W$ are the cyclotomic fields $\mathbb{Q}(\zeta_{m_3})$, $\mathbb{Q}_(\zeta_{m_2})$, and $\mathbb{Q}(\zeta_{m_1})$ respectively, where $m_1 \mid m_2 \mid m_3$ so that $U/V/W$ is a tower of ...
3
votes
1answer
177 views

Reference for constructing tensor products of finitely presented functors

I need references related to the construction of tensor product between functors Let $k$ be a commutative ring, $C$ a small $k$-linear category and $A$ cocomplete abelian category. Let $A^C$ denote ...
1
vote
2answers
101 views

Irreducibility of tuples of matrices

I am interested in tuples of real or complex $d\times d$ matrices which are irreducible in the sense that their constituent matrices do not admit a common invariant linear subspace whose dimension is ...
9
votes
1answer
215 views

Enriching categories and equivalences

Let $\mathcal{C}$ and $\mathcal{D}$ be two equivalent categories. Furthermore, assume $\mathcal{C}$ is enriched over a monoidal category $(\mathcal{M}, \otimes)$. Can one use the equivalence to ...
5
votes
1answer
210 views

Plethysm of $S^3(S^2V)$ as $\mathfrak{sl}_3(\mathbb{C})$-module

I have asked this question in MSE before, but have not got any answer. So here I am asking it again with some more detail. I believe that the following sequence of ...
4
votes
1answer
75 views

second dual of minimal tensor products of $C^*$-algebras

Let $A$ be a unital $C^*$-algebras and $K(H)$ is $C^*$-algebras of compact operators on separable Hilbert space $H$. Is it true that $(A \otimes K(H))^{**}= A^{**} \overline{\otimes}B(H)$?
2
votes
1answer
179 views

Kunneth formula of Cartesian product modulo orders of coordinates

Let $X$ be a topological space and $F$ a field. Let the $n$-th permutation group $\Sigma_n$ act on $$ \prod_n X $$ by $$ \sigma(x_1,\cdots,x_n)=(x_{\sigma(1)},\cdots,x_{\sigma(n)}), \sigma\in ...
0
votes
1answer
65 views

Ideal in projective tensor product of Banach algebras [closed]

Let $A,B$ be Banach algebras and $A\hat{\otimes}B$ be projective tensor product of them. Let $S$ be an ideal of $A\hat{\otimes}B$. Are there ideals $I$ of $A$ and $J$ of $B$ such that ...
3
votes
0answers
128 views

Annihilators of elements in symmetric algebras

Let $M$ be a module over a commutative ring, and $S(M)$ its symmetric algebra. What elements of $S(M)$ annihilate a given element $m\in M$ ? ($M$ is considered as a submodule of $S(M)$.)
5
votes
0answers
224 views

C$^*$-algebras isomorphic after tensoring

From the negative answer to this question we know that C$^*$-algebras that are isomorphic after tensoring with $M_n$ for all $n\geq 2$ need not be isomorphic. So what happens when we strengthen this? ...
1
vote
1answer
92 views

One question about the tensor product of $L^1(G)$ and a Banach space $A$

We know that the tensor product of $L^1(G)$ and a Banach space $A$ is isometric to $L^1(G, A)$, the space of all Bochner-integrable $A$-valued functions on a locally compact group $G$. I am looking ...
3
votes
1answer
432 views

Comparing norms on tensor products of matrices

Given a Hilbert space $H$, let $S_1(H)$ denote the space of trace-class operators on $H$, with the trace-class norm or Schatten 1-norm. That is $$ \Vert T \Vert_1 = \sum_{j\geq 1} |s_j| $$ where ...
2
votes
1answer
112 views

Symplectic form on the third symmetric power of a plane

Let $V$ a vector space of dimension $2$ over a field $k$ of characteristic different from $2$ and $3$. Let $S^{3}V$ the third symmetric power of $V$. How to construct a symplectic form on $S^{3}V$ ...
2
votes
0answers
77 views

Tensor product of bounded analytic functions

I asked this question on math.SE, but couldn't get an answer. Let $H^\infty(\mathbb{D})$ denote the set of functions holomorphic and bounded on $\mathbb{D} = \{z \in \mathbb{C}: |z| < 1\}$. ...
2
votes
2answers
173 views

Permutation covering of a $G$-lattice

Let $G$ be a finite group. By a $G$-lattice we mean a finitely generated free abelian group $L$ with an action of $G$. We say that $L$ is a permutation $G$-lattice if $L$ has a ${{\mathbf{Z}}}$-basis ...
4
votes
1answer
331 views

Irreducibility of the tensor product of two finite-dimensional irreducible group representations

Let $k$ be an algebraically closed field of characteristic 0, let $G$ be any group and $N\unlhd G$ a normal subgroup. Let $U$ be a finite-dimensional and irreducible $kG$-module, such that $U$ is also ...
3
votes
0answers
330 views

3 possible tensors in 2-categories?

let $\mathcal{A}$ be a 2-category, consider: $$ \mathcal{A}(W \otimes_i A, X) \simeq_i \mathcal{C}at(W, \mathcal{A}(A, X)) \;\;\; i = 1, \: 2, \: 3. $$ where $W$ is a category, and $A$, ...
0
votes
2answers
168 views

Regularity of a tensor product

Let $A \subseteq B$ and $A \subseteq C$ be commutative noetherian domains. Assume that $A$ and $C$ are regular rings (=every localization at a maximal ideal is a regular local ring). Assume that $B$ ...
0
votes
1answer
155 views

When every module is a scalar extension?

Let $A \subseteq B$ be commutative noetherian domains. Of course, if $M$ is an $A$-module, then $M \otimes_A B$ is a $B$-module. I am curious to know if there exist additional conditions on $A$ and ...
6
votes
0answers
184 views

A “slice-map” type problem for symmetric tensors in the square of a nuclear C*-algebra

Throughout: let $\otimes$ denote the minimal (i.e. spatial) $\newcommand{\Cst}{{\rm C}^*}\Cst$-tensor product of two $\Cst$-algebras. Let $B$ be a unital, nuclear $\Cst$-algebra and let $A\subset B$ ...
1
vote
0answers
198 views

Does the Boardman-Vogt tensor product of operads commute with their W-construction

I have absolutely no idea whether this is true or not but it could well be useful for me in the future if it is. If we have topological operads $\mathcal{P}$ and $\mathcal{Q}$ and we let $W$ denote ...
9
votes
0answers
124 views

When does Hochschild homology commute with infinite products?

Let $A$ be an associative algebra. Its zeroth Hochschild homology $\mathrm{HH}_0(A)$ is the cokernel of the linear map $A^{\wedge 2} \to A$, $a \wedge b \mapsto ab - ba$. I.e. you quotient the ...
0
votes
0answers
156 views

Symmetric kernel of tensor product

Let $V,W$ be two vector spaces, and let $L_i:V\rightarrow W$, $i=1,\ldots,n$ be $n$ linear maps with disjoint kernels $K_i$ of dimension $1$. Consider the tensor product of these maps $L_1\otimes ...
4
votes
2answers
255 views

Topologies for which $\mathcal{M}(X)\otimes \mathcal{M}(Y)$ is dense in $\mathcal{M}(X\times Y)$

Are there complete TVS topologies for which $\mathcal{M}(X)\otimes \mathcal{M}(Y)$ is dense in $\mathcal{M}(X\times Y)$ This question is strongly linked to is the space of all borel measures on ...
5
votes
1answer
204 views

When is the direct product of two graph cores itself a core?

A graph homomorphism $f$ is a function $f : V(X) \to V(Y)$ such that if $uv \in E(X)$, then $f(u)f(v) \in E(Y)$. If such an $f$ exists, write $X \to Y$. $X$ and $Y$ are hom-equivalent if $X \to Y$ and ...
6
votes
1answer
254 views

Making the identification $\tau M\approx TM\oplus (TM\odot TM)$

Given a smooth manifold $M$, there is a vector bundle over $M$, denoted $\tau M$, known as the second-order tangent bundle. The fiber $\tau_mM$ at $m\in M$ is the collection of linear operators ...
1
vote
1answer
200 views

Necessary and sufficient condition for $can : A^X\otimes_A A^Y\rightarrow A^{X\times Y}$ to be an embedding

The two sets are, of course, supposed infinite. This question is related to that one Commutation of tensor products with inverse limits in a specific case where it received a (partial) answer ($A$ ...
0
votes
1answer
206 views

Commutation of tensor products with inverse limits in a specific case

For $X,Y$ sets, let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well, a bi-module but my question is - at first - concerning commutative ...
1
vote
1answer
157 views

Annihilator of tensor product when $R$ is domain

Let $R$ be a Noetherian domain and $M$ and $N$ be two faithful $R$-modules. Is it true that $\operatorname{Ann}_R(M\otimes_R N)=0$?
1
vote
0answers
77 views

A reasonable framework to study properties of operator $A \mapsto KAK$ on Banach space

Let $K$ be a continuous linear operator on $C[0,1]$ (more, precisely, it is a linear integral operator). Then $K$ defines a continous linear operator $\widehat K$ on $\mathcal L(C[0,1])$ by the rule ...
1
vote
1answer
137 views

The tensor product of admissible morphisms of semi-normed modules over a normed ring is an admissible morphism (V. G. Berkovich)

Disclaimer : I found here http://mathoverflow.net/editing-help in the spoilers paragraph that putting >! would hide following things, which was a way for me to alleviate my question's presentation by ...
4
votes
1answer
255 views

Can the method of small moments prove a bound on the norms of random trilinear forms?

If $F(v_1,\dots,v_k)$ is a $k$-linear form on $\mathbb R^n$, the norm I want to consider is $$ ||F|| = \sup \frac{ F(v_1,\dots, v_k)}{\prod_{i=1}^k \left|\left|v_i\right|\right|} $$ where the vector ...
0
votes
2answers
122 views

Tensor product-definition-balanced versus bilinear maps

When defining tensor products $M\otimes_R N$ over a commutative ring $R$ one can use a universal property with respect to bilinear maps $M\times N\rightarrow P$, for any $R$-module $P$. On the other ...
6
votes
2answers
254 views

Exactness of an additive left Kan extension

Let $\phi:R\to S$ be a flat ring homomorphism and consider the induced adjoint pair $$\phi_!:R-Mod\rightleftarrows S-Mod:\phi^*,$$ where $\phi_!=(S\otimes_R -)$. The right adjoint $\phi^*$ is easily ...
0
votes
1answer
125 views

Complementation in tensor products

This question, however looks innocent, looks non-trivial to me. Suppose that $X$ and $Y$ are Banach spaces and let $\alpha$ be any reasonable cross-norm on $X\otimes Y$. Reasonable means that ...
0
votes
1answer
447 views

Why does an infinite tensor product depend on some vectors for operator algebras?

I have read that, in the definition of the infinite tensor product of operator algebras such as ${\rm C}^\ast$-algebras and ${\rm W}^\ast$-algebras, every factor in the product is associated with a ...
6
votes
1answer
257 views

Tensor product of certain Sobolev spaces on non-compact manifolds

Let $M$ be a non-compact Riemannian manifold of bounded geometry (i.e., its injectivity radius is uniformly positive and the curvature tensor and all its covariant derivatives are bounded in ...
2
votes
0answers
124 views

Tensor product of commutators vs. commutator in a tensor product

Let $R$ be a (noetherian) commutative ring, and let $V$ and $W$ be finitely generated free $R$-modules. Let $X \subseteq \mathrm{End}_R(V)$ and $Y \subseteq \mathrm{End}_R(W)$ be finite subsets, and ...
0
votes
4answers
213 views

about the structure of components of tensor product if more than one bipartite graph is taken

I was reading about tensor product of graphs. We know that if we take tensor product of n graphs and want this product to be a connected graph then at most one graph should be bipartite. In the book ...
2
votes
0answers
135 views

Inductive and projective tensor product

Does anyone know if there is a characterization of the spaces on which the inductive tensor product and the projective tensor product are the same ? This is the same as asking every separately ...
9
votes
2answers
431 views

Continuity of the product map

Let $A$ be a $C^*$-algebra. Is it possible to characterize $A$ for which the product map defined by $$\sum\limits_{i=1}^n a_i\otimes b_i \mapsto \sum\limits_{i=1}^n a_i b_i$$ is continuous with ...
2
votes
0answers
95 views

Smooth function over a manifold into an algebra

Let $M$ be a manifold and $A$ a $*$-algebra. Does is hold that $$C^{\infty}(M,A) \cong C^{\infty}(M) \otimes A$$ where the RHS means that you take smooth functions which map into $A$. If this holds, ...
5
votes
0answers
161 views

Is there a tensor norm that preserves Rosenthal Banach spaces?

By a Rosenthal Banach space I mean one that does not contain an isomorphic copy of $\ell_1$. Is there a tensor norm $\alpha$ such that the Banach tensor product $E\otimes_\alpha F$ is Rosenthal if $E$ ...
7
votes
1answer
291 views

Infinite tensor product of states

Tensor products of finite number of different objects are always well described in the literature. However, the situation of infinite tensor products seems to be much tougher. Even in the simplest ...