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This is the most complete treatment I could come up with. Let $k \subseteq K^{\operatorname{sep}} \subseteq K^{\operatorname{alg}} \subseteq K$ and $k \subseteq L^{\operatorname{sep}} \subseteq L^{\operatorname{alg}} \subseteq L$ be the separable algebraic and algebraic closures of $k$ in $K$ and $L$. The result is the following.
Theorem. Let $k \subseteq K$...

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Here is a self-contained argument. First, as Jeremy Rickard observes, $K \otimes K \cong K \otimes_k K$, where $k$ is the prime subfield of $K$ (so $\mathbb{Q}$ if $K$ has characteristic zero and $\mathbb{F}_p$ if $K$ has characteristic $p$). If $K \otimes_k K$ is a field, then as Denis Nardin observes, the multiplication map
$$K \otimes_k K \xrightarrow{m} ...

answered Feb 15 '16 at 4:26

Qiaochu Yuan

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20

I already wrote this in the comments but I think this might be worth of an answer. I think we can classify all fields $K$ such that $K\otimes K$ is a field.
Claim If $K$ is a field such that $K\otimes_\mathbb{Z}K$ is a field then the multiplication map $K\otimes_\mathbb{Z} K\to K$ is an isomorphism
In fact the multiplication map is always a surjection ...

16

The answer to the question posed in the title of your post is yes, the tensor product of chain complexes is a Day convolution product. The important thing to note is that, to define a Day convolution monoidal structure on the $\mathcal{V}$-enriched functor category $[\mathcal{C},\mathcal{V}]$ (where $\mathcal{V}$ is a complete and cocomplete symmetric ...

answered May 4 '20 at 10:49

Alexander Campbell

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I think both can be proved without choice, essentially because, in both cases, whenever you're tempted to choose a basis, you can manage with a little care to get by with a basis of a finite dimensional subspace.
For (2), if there's a linear dependence between the $v_i\otimes w_j$ then it involves only finitely many $v_i$ and $w_j$. Also, the linear ...

answered Mar 10 '19 at 10:11

Jeremy Rickard

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12

Recall that $\mathcal Z$ is a weak $P$-point if it is not in the closure of any countable subset of $\omega^* \setminus \{\mathcal Z\}$. A weak $P$-point is never the tensor product of two non-principal ultrafilters.
To see this, suppose $\mathcal U$ and $\mathcal V$ are two non-principal ultrafilters on $\omega$, and let $\mathcal Z = \mathcal U \times \...

12

Pisier https://arxiv.org/abs/1908.02705 very recently constructed a non-nuclear $C^\ast$-algebra $A$ with the weak expectation property (WEP) and the local lifting property (LLP). By a celebrated result of Kirchberg (see Corollary 13.2.5 in Brown and Ozawa's book) it follows that if $B$ has WEP and $C$ has LLP, then $B\otimes_{\max{}} C = B\otimes_{\min{}} C$...

12

Provided at least one of $M$ and $N$ is locally compact, the $\infty$-topos $\mathrm{Sh}(M \times N)$ is the product of $\mathrm{Sh}(M)$ and $\mathrm{Sh}(N)$ in $\mathrm{RTop}$. This is HTT 7.3.1.11.
Products in $\mathrm{RTop}$ can be computed as tensor products in $\mathrm{Pr^L}$. This is HA Example 4.8.1.19.
HA doesn't include a complete proof of the ...

11

$$ \mathrm{End}_{kG}(U \otimes_k V) = \left(\mathrm{End}_k(U \otimes_k V)^N\right)^{G/N} = \mathrm{End}_k(V)^{G/N} = k. $$
The second equality holds because $\mathrm{End}_{kN}(U) = k$ (thanks to the assumption that $k$ is algebraically closed).
EDIT: Using Chevalley's theorem is probably overkill. Let $k$ be an arbitrary commutative field, assume that $U$ ...

10

Why can't you just take the product measure induced by the uniform measure on $\{0,1\}$ (or indeed by any other nontrivial measure on this two-element set)? I suppose my question really is: What exactly do you mean by non-trivial?

answered Nov 21 '13 at 19:57

Andreas Blass

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10

I think the answer is: It's always continuous if $A$ is subhomogeneous and never continuous otherwise(min or max).
First notice that if the product map is continuous for min, then it's continuous for max because we can factor the product map as $A\otimes_{max}A\rightarrow A\otimes_{min}A\rightarrow A$ where the first map is the canonical quotient and the ...

10

This answer provides a scheme how to construct a constructive proof, though I'm still working to actually explicitly extract the constructive proof, so please don't accept the answer just yet. (Update: See below.) We'll prove the following statement:
Let $R$ be a reduced ring. Let $A$ be a finitely generated $R$-module and let $B$ be an arbitrary $R$-...

10

Many introductory books on quantum information theory go over the linear algebraic tools necessary to study the topic, including the tensor product (since it indeed models quantum entanglement). Taking the tensor product of two or more quantum states (pieces of quantum information) is analogous to forming a bitstring out of two or more bits (pieces of ...

answered Mar 5 '20 at 14:00

Nathaniel Johnston

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10

Planar partitions with no singletons works. You need to pick for each $n>1$ some map with certain properties. One way to do this is to just fix a preferred trivalent tree of each size and interpret each vertex as a cross product. For example, one arbitrary choice gives the map $V^{\otimes n} \rightarrow \mathbb{C}$ given by $$v_1 \otimes \ldots \otimes ...

10

Here's a sketch of the proof. I encourage you to fill in the details yourself. The definition of $V_K$ is $V_K=H^0(G_{\overline K/K},V)$. The key part of the proof is to show that $V$ has a $\overline{K}$ basis consisting of vectors in $V_K$. To find such a basis, start with an arbitrary basis $v_1,\ldots,v_n\in V$. Then each $\sigma\in G_{\overline K/K}$ ...

9

If you're working at the level of Hilbert spaces, I think the more usual procedure nowadays is to fix a distinguished unit vector $u_i$ in each Hilbert space $H_i$, then define the tensor product to be the Hilbert space generated by all elementary tensors $\bigotimes v_i$ such that $v_i \in H_i$ for all $i$ and $v_i = u_i$ for all but finitely many $i$. This ...

9

I asked David Sherman this question and this was his response:
In the Kuratowski setup, if you omit complementation, you get 7 elements. Adding complementation gives you a disjoint upside-down copy.
In the Grothendieck setup, if you omit duality, you get 8 elements (two five-element chains with common inf and sup, like an O). Adding duality gives you an ...

9

I think most category theorists would answer "yes, obviously", and not bother to write down a proof. But presumably that isn't sufficiently convincing, since you ask the question, so let me try to make it a bit more explicit with some big words. (-:
An $M$-enriched category with set of objects $A$ is equivalently a lax functor $A_{ch}\to B M$, where $A_{...

answered Oct 14 '15 at 4:53

Mike Shulman

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9

Look at $M=S^7$. This has 28 smooth structures, and their tangent bundles are all trivial - compare Parallelizability of the Milnor's exotic spheres in dimension 7 . Thus their tensor algebras do not carry more information about the smooth structure than $C^\infty(M)$.
On the other hand, for a compact smooth manifold, the algebra of smooth functions is ...

9

Here's an explicit example. Let $R=\mathbb{Z}[\sqrt{2}]$, let $X=\mathbb{P}^1_R$, and let $Y$ be the smooth projective conic defined by the equation $$(2-\sqrt{2})x^2+y^2+(2-\sqrt{2})z^2+xy+yz+(3-2\sqrt{2})xz=0.$$
I claim this conic has good reduction everywhere; as smooth conics over the ring of integers of a $p$-adic field are always split, this implies ...

9

Your $\vee$ is essentially multiplication of polynomials. The variety of tensors $x_1 \vee \dotsb \vee x_m$ corresponds to polynomials that factor as products of linear factors. Points of the (projective) variety correspond to hyperplane arrangements. Dually, they correspond to cycles of $m$ points (in the dual projective space).
This variety has various ...

8

$2^{\mathbb{R}}$, being a product of compact Hausdorff groups, is a compact Hausdorff group, so it has a normalized Haar measure ("flipping uncountably many coins").

answered Nov 22 '13 at 0:34

Qiaochu Yuan

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8

Looks like nfdc23 has explanations for (a),(b), and (c).
But: indeed, primes of $F_{k'}$ are not in general minimal if $F/k$ is not algebraic. Let $k'=k(x)$ and $F=k(y)$ be transcendental extensions of $k$.
Then $F_{k'}$ identifies with a subalgebra of the field $k(x,y)$, hence is an integral domain [assertion (c)] - i.e. $\{0\}$ is prime.
There is a ...

8

Taking your last question first: there is an accessible discussion of the link between the AP and the injectivity of the comparison map "projective tensor product to injective tensor product" in Section 4.1 of
R. Ryan, Introduction to Tensor Products of Banach Spaces, Springer Monographs in Mathematics, 2002.
In particular, it seems that one can extract ...

8

The first thing that comes to mind is the symmetric algebra functor
$$A \mapsto Sym_{\mathbb Z}(A) = \oplus_{n \in \mathbb N} Sym^n_{\mathbb Z}(A) = \mathbb Z \oplus A \oplus (A\otimes A/ \Sigma_2) \oplus \dots $$
This may be regarded as a functor from abelian groups to commutative rings (it is left adjoint to the forgetful functor); or by composing with ...

8

The answer is no. Bourgain and Pisier have given a counterexample (A construction of $\mathcal{L}_\infty$-spaces and related Banach spaces. Bol. Soc. Bras. Mat. 14, No. 2, 109-123 (1983). See Zbl 0586.46011 https://zbmath.org/?q=an%3A0586.46011 ).

8

Nate's suggestion on math.SE works. We'll show that if $A = k[x, \partial_x]$ and $B = k[y, \partial_y]$ are both taken to be the Weyl algebra, then the module over $A_2 = A \otimes B \cong k[x, \partial_x, y, \partial_y]$ generated by $e^{xy}$ is 1) simple and 2) not a tensor product of simple modules of $A$ and $B$.
Explicitly this module $M$ consists of ...

answered Sep 10 '20 at 0:04

Qiaochu Yuan

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8

According to John Aldrich's list of "Earliest Uses of Symbols for Matrices and Vectors", the notation $\times$ for direct product (as well as the name itself) goes back to Wedderburn's 1934 Lectures on Matrices (page 74).
Some further search gave a much earlier source, Hurwitz's 1894 Zur Invariantentheorie
I still have to track down the step from ...

answered Dec 2 '20 at 10:00

Carlo Beenakker

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8

The answer is NO, if you count $0$ as a unital Banach algebra. Otherwise, it's YES. Let $e \in A \hat\otimes B$ be a unit and take $b_0\in B$ and $g\in B^*$ with $g(b_0)=1$. Let $g\cdot b_0\in B^*$ be defined by $g\cdot b_0\colon x\mapsto g(b_0x)$. Then $e_A:=(\mathrm{id}\otimes (g\cdot b_0))(e) \in A$ is a left unit for $A$.
Indeed, $$ae_A=(\mathrm{id}\...

7

In case anyone is still interested in this question:
There is a classification of polynomial exponential functors on the category $\mathcal{V}$ of finite-dimensional inner product spaces in terms of involutive $R$-matrices (i.e. involutive solutions to the Yang-Baxter equation). Basics about polynomial functors can be found in the book "Symmetric Functions ...

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