34
votes

### When is the tensor product of two fields a field?

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^{\...

Community wiki

22
votes

### Tensor product of fields over integers

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 $\...

20
votes

Accepted

### Tensor product of fields over integers

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\...

19
votes

Accepted

### Is the tensor product of chain complexes a Day convolution?

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 ...

13
votes

Accepted

### Axiom of choice and algebraic tensor product

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 ...

12
votes

Accepted

### Non-tensor-representable ultrafilters on $\omega$

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-...

12
votes

Accepted

### A non nuclear $C^*$ algebra $A$ for which the algebraic tensor product $A\otimes A$ admits a unique $C^*$ norm

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 ...

12
votes

Accepted

### Taking the category of sheaves is symmetric monoidal

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.
...

11
votes

Accepted

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

$$ \mathrm{End}_{kG}(U \otimes_k V) = \left(\mathrm{End}_k(U \otimes_k V)^N\right)^{G/N} = \mathrm{End}_k(V)^{G/N} = k. $$
The second equality holds because $\mathrm{End}_{kN}(U) = k$ (thanks to the ...

11
votes

Accepted

### Constructive proof that a kernel consists of nilpotent elements

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:...

11
votes

### Linear algebra underlying quantum entanglement?

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). ...

11
votes

### Proof of $V\cong \overline{K} \otimes_{K} V_K$ using $H^1(G_{\overline{K}/K},\operatorname{GL}_n(K))＝0$

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{...

10
votes

Accepted

### Recovering a smooth manifold from its tensor fields

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 ...

10
votes

Accepted

### Basis of invariant tensors of rank n in three dimensions

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 ...

10
votes

### Clebsch–Gordan decomposition formula for algebraic groups

Let $G$ denote the group, and suppose one has an enumeration of its irreducible representations $V_{\lambda}$ by some combinatorial objects $\lambda$, like integers for $SU_2$ (or finite dimensional ...

9
votes

Accepted

### Enriching categories and equivalences

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 ...

9
votes

### Functors on the category of abelian groups which satisfy $F(G\times H) \cong F(G)\otimes_{\mathbb{Z}} F(H)$

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) \...

Community wiki

9
votes

Accepted

### Varieties with everywhere good reduction that are isomorphic over every completion have isomorphic generic fibers

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\...

9
votes

Accepted

### A different notion of a decomposable symmetric tensor (besides Veronese)

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 (...

9
votes

Accepted

### Defining the abstract tensor product of W*-algebras via a universal property

You can see this is false by taking $N = \mathbb{C}$. Then, given a von Neumann algebra $M$, you are asking for a von Neumann algebra $\widetilde{M}$ and a weak* dense embedding $\iota: M \to \...

9
votes

### Injection vs. bijection between $V^\star \otimes V^\star$ and $(V \otimes V)^\star$

Write $V=k^{(I)}$ (this is always the case, up to isomorphism, for some set $I$, namely take $I$ to be a basis subset).
So $V^*=K^I$, $V\otimes V=k^{(I\times I)}$ and the embedding of $V^I\otimes V^I$ ...

8
votes

Accepted

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

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 ...

8
votes

### Approximation property counterexamples? (Also: relation to tensor products)

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 ...

8
votes

Accepted

### Containment of $c_0$ in projective tensor products

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....

8
votes

Accepted

### Infinite dimensional irreducible representations of a tensor product

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, \...

8
votes

### Origin of the symbol for the tensor product

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 ...

8
votes

### If $A\hat\otimes B$ has identity then so are $A$ and $B$

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 ...

7
votes

Accepted

### exponential functors on finite dimensional complex vector spaces

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 ...

7
votes

Accepted

### Counting with tensor products

You are looking at the vector of coefficients of the polynomial
$$\sum_{\epsilon}\prod_{i=1}^{m+2}(1-\epsilon_i x_i)$$
where $\epsilon$ runs over all choices of signs $\pm$ provided there are exactly ...

7
votes

Accepted

### Tensor product of field extensions

In general, if $A$ is a field of characteristic $0$ and $B/A, C/A$ are two finite extensions of $A$, then the tensor product $B \otimes_A C$ is isomorphic to the product (i.e. direct product of $K$-...

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