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^{\...
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 $\...
- 112k
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\...
- 16k
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 ...
- 4,945
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 ...
- 31.7k
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-...
- 16.8k
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 ...
- 2,271
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.
...
- 24.5k
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 ...
- 126
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:...
- 3,062
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). ...
- 5,413
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{...
- 44k
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 ...
- 4,534
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 ...
- 27.3k
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 ...
- 20.9k
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 ...
- 62.7k
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
$$A \mapsto Sym_{\mathbb Z}(A) = \oplus_{n \in \mathbb N} Sym^n_{\mathbb Z}(A) = \mathbb Z \oplus A \oplus (A\otimes A/ \Sigma_2) \...
9
votes
Accepted
Varieties with everywhere good reduction that are isomorphic over every completion have isomorphic generic fibers
Here's an explicit example. Let $R=\mathbb{Z}[\sqrt{2}]$, let $X=\mathbb{P}^1_R$, and let $Y$ be the smooth projective conic defined by the equation $$(2-\sqrt{2})x^2+y^2+(2-\sqrt{2})z^2+xy+yz+(3-2\...
- 21.7k
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 (...
- 5,450
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 \...
- 40.5k
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 $V^I\otimes V^I$ ...
- 56k
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 ...
- 3,126
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 ...
- 24.8k
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....
- 1,736
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, \...
- 112k
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 ...
- 163k
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 ...
- 9,446
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 ...
- 7,377
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 ...
- 83.4k
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 $K$-...
- 3,140
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