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6 votes

Two definitions of antisymmetrization of a tensor?

Your convention for the factor in front of the antisymmetrization determines a corresponding convention for the factor in front of the exterior product. The factor is uniquely determined by the ...
Qiaochu Yuan's user avatar
3 votes

Two definitions of antisymmetrization of a tensor?

See my answer here. In short (and also to expand on it a bit), using you definition of $\mathscr A$ with the $1/k!$ out front lets us define a wedge product such that $$ v_1\wedge\dotsb\wedge v_k := ...
Nicholas Todoroff's user avatar
3 votes
Accepted

Operator inner product

This is not true. Indeed, here is a counterexample: $$A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \; v = e_1 \otimes e_2, \; \langle v, A \otimes Av \rangle = -1$$ Here is a suggested ...
David Gao's user avatar
  • 9,745
3 votes
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What is the relation between the tensor of two $R$-modules over different rings.

A sufficient condition is that the map $\mathbb{Z} \to R$ is an epimorphism; this includes the case of quotients but is more general since it also includes localizations and some other stranger things....
Qiaochu Yuan's user avatar
3 votes
Accepted

Definition of tensor product seems to contradict universal property

Tensor products A canonical definition of a tensor product is by the universal property: The tensor product of the vector spaces $U$, $V$ is a vector space $U \otimes V$, together with a bilinear ...
Adayah's user avatar
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2 votes
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Why is the set of decomposable tensors closed under any norm-induced topology?

On finite-dimensional spaces, all norms are equivalent, so we might as well pretend $V_i$ are all inner product spaces and the norm on $V_1 \otimes \cdots \otimes V_k$ is induced by the usual inner ...
David Gao's user avatar
  • 9,745
2 votes
Accepted

For two irreducible modules $V$ and $W$, $f : V\to W$ is a $G$-module isomorphism $\iff \text{span}(f)\subset V^*\otimes W$ is trivial

Perhaps it is better to identify $V^* \otimes W$ with $Hom(V,W)$. The action of $G$ on $Hom(V,W)$ is then seen to be $$ g \cdot f := (g \cdot f)(v) = g \cdot f(g^{-1} \cdot v)$$ Recall now the ...
Guster's user avatar
  • 401
1 vote

Grounding the concept of a Free Vector space of the cartesian product of two vector spaces

Your comment is correct, you can start with a $2$-element basis $\{v_1,v_2\}$ for $V$ and a $3$-element basis $\{w_1,w_2,w_3\}$ for $W$ and from them you can produce a $6$ element basis for $V \otimes ...
Lee Mosher's user avatar
  • 124k
1 vote

Two definitions of antisymmetrization of a tensor?

Rewritten shorter but more rigorous answer: You have to distinguish between a vector space $V$ and its dual $V^*$, which is a derived object. $V^*\otimes V^*$ is easier to define than $V\otimes V$. It ...
Deane's user avatar
  • 8,502
1 vote
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Constructing a basis for a tensor product

If the pair $(V\otimes W,\otimes)$ is a tensor product of vector spaces $V$ and $W$ with tensor product map $\otimes:V\times W\to V\otimes W$, then the pair satisfies the universal property: For any ...
blargoner's user avatar
  • 3,291
1 vote
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Do tensor products remain unchanged along ring epimorphisms?

Yes. The easiest way to see this is to write everything in terms of tensor products over $S$, which means $M \otimes_R N$ is rewritten $$M \otimes_S \left( S \otimes_R S \right) \otimes_S N$$ and we ...
Qiaochu Yuan's user avatar
1 vote
Accepted

Double Trace of the tensor product of the metric tensor with vector fields.

The notation you are using partly is not really well defined. There are (at least) two possible definitions of what $tr[g\otimes X\otimes Y]$ should mean, namely either $g(X,\ )\otimes Y$ or $g(\ ,Y)\...
Andreas Cap's user avatar
  • 21.2k
1 vote

Discrete spectrum of $A \otimes 1+ 1 \otimes B$

Let's complete the hints of Christian Remling to an actual answer. As shown here, the spectral measure $E$ of $A\otimes 1+1\otimes B$ satisfies $$ E(S)=\int_{\sigma(B)}\int_{\sigma(A)}1_S(\lambda+\mu)\...
MaoWao's user avatar
  • 15.4k

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