5 votes
Accepted

$ SO_4(\mathbb{R}) $ and $ SU_2 \otimes SU_2 $ subgroups of $ SU_4 $

Yes, $SU(2)\otimes SU(2)$ is conjugate to the usual $SO(4)$ in $SU(4)$. There's probably a direct way to see it, but here's some theory. Suppose $G = G_1\times G_2$ is a product of compact Lie groups....
  • 46.8k
4 votes
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Trying to compute $\operatorname{kernel}$ of the map $\pi\otimes \rho: A \otimes^{\text{min}} B \to A/I \otimes^{\text{min}} B/J$

The following is a deep result, recorded as corollary 9.4.6 in Brown Ozawa: If $A$ is an exact $C^*$-algebra and $B$ is any $C^*$-algebra and $I\subset A\otimes B$ is an ideal, then $$I=\overline{\...
3 votes

Understanding symmetric tensor products

The symmetric tensors are all symmetrized tensor products of vectors: $$ H^s_n = \left\{\sum_{\sigma \in S_n}v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(n)} \;:\; v_1,\dotsc,v_n \in H\right\}. $$ It ...
3 votes
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Prove a map of tensor product is linear.

As hinted in the comments, you'll get this for free with a construction of $\tau$ that ensures it is actually well-defined. Not every tensor in $L\otimes M$ is decomposable in the form $\alpha\otimes\...
  • 1,678
3 votes
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Relationship between a tensor and the tensor product

Short answer: To a mathematician, a tensor is just an element of a tensor product. As you ask in the comments, it is correct to say that a tensor is to a tensor product what a vector is to a vector ...
2 votes

$ SO_4(\mathbb{R}) $ and $ SU_2 \otimes SU_2 $ subgroups of $ SU_4 $

The $\mathbb{C}$-vector space isomorphism $\mathbb{C}^2\otimes\mathbb{C}^2\cong\mathbb{C}(2)$ (the latter being the space of $2\times2$ complex matrices) given by $u\otimes v\mapsto uv^T$ allows us to ...
  • 16.9k
2 votes
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Notation in defining the abstract tensor product

We have two different kinds of vector spaces, the free product and the actual product. Let $V_1,\ldots,V_n$ be a collection of $n$ (real) vector spaces. We can form the product vector space $$ V_1\...
  • 505
2 votes
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Equivalent definitions of the tensor product.

Your suspicion is correct: the definition given in your differential geometry class works for finitely generated free modules but not in general. There is a more general definition found in many ...
2 votes

Definitions - Tensor Products and Tensors

Let me try to answer question (ii). If $V$ is a vector space over a field $F$, then the space of $(p,q)$-tensors over $V$ is the space $$T^p_q(V)=V^{\otimes p} \otimes (V^*)^{\otimes q}.$$ A $(p,q)$-...
2 votes
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length of longest shortest path in the tensor/kronecker square of a graph G?

Here's a lower bound showing that it may be at least $\lceil \frac{n^2}{2}\rceil$. Let $G$ be the graph with vertices $1, 2, 3, \dots, n$ and edges $1 \to 2 \to 3 \to \dots \to n$ together with $n \to ...
2 votes
Accepted

Question about defining maps on Tensor Product Space

$v\otimes w$ is the definition of $\phi(v,w)$. The tensor product is a vector space $V\otimes_FW$ together with a bilinear map $\phi:V\times W\to V\otimes_FW$ which satisfy the following universal ...
  • 33.6k
2 votes
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Understanding symmetric tensor products

A (small) example to get the feel of these objects: $$ 2 \, e_3 \otimes e_{17} - 5 \, e_4 \otimes e_4 $$ is an arbitrary element of the two-fold tensor product space $H_2$. Here the multi-index $(i_1,...
  • 18.6k
1 vote
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Confusion regarding the definition of tensor product of modules.

Based on your post, I'm going to assume that $R$ is commutative. A tensor product of the $R$-modules $M$ and $N$ is a pair $(T,\varphi)$ where $T$ is an $R$-module and $\varphi:M\times N\to T$ is a ...
  • 1,678
1 vote

Notation in defining the abstract tensor product

To illustrate with a simple example, let's take $n=2$ and $V_1=V_2=\mathbb{R}$. Then $F=\mathcal{F}(\mathbb{R}\times\mathbb{R})$ is the space of formal finite linear combinations of pairs of real ...
  • 1,678
1 vote
Accepted

change of rings for faithfully flat modules-Boubaki's proof

For any $A$-module $M$ and any ideal $I$ of $A$, the natural map $M \otimes_{A} A/I \to M/IM$ sending the simple tensor $m \otimes [a]$ to $[am]$ is an isomorphism of $A$-modules. If $\rho \colon A \...
1 vote
Accepted

Product of projection maps

It is not exactly tensor product, but composition of tensor product and diagonal map. In general for $f:X\to Y$ and $g:A\to B$ we define $$ f\times g : X\times A \to Y\times B,\quad (f\times g)(x, a) =...
  • 1,090
1 vote
Accepted

A scalar field satisfying the Laplace's Equation is zero everywhere

I'm going to assume that $S$ is smooth closed surface that is embedded into $\mathbb R^3$. The smoothness assumption can be relaxed to $C^1$. Also, I think the assumption that $S$ is embedded is ...
  • 4,296
1 vote

Is this how to prove the two-sided ideal is spanned by $v_1\otimes\dots\otimes v_k$ where the $v_1,\dots,v_k$ are linearly dependent?

Let $I_n$ be the subspace of $V^{\otimes n}$ generated by elementary tensors $v_1\otimes\cdots\otimes v_n$ in which two consecutive factors are equal, and let $f:V\times\cdots\times V\to V^{\otimes n}/...
1 vote
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Decomposition of an operator $x\in B(H\otimes K)$ as a sum $x=\sum_{i,j} x_{ij}\otimes E_{ij}$.

You have the identity$\def\e{\epsilon}$ $$\tag1 \sum_i\pi_i(\xi\otimes\e_j)\otimes \e_i=\xi\otimes\e_j. $$ Let $\tilde y=\sum_{\ell\in G_0} \eta_\ell\otimes\e_\ell$, the sum over some finite set $G_0$....

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