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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 ...
• 431k
3 votes

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 $$v_1\wedge\dotsb\wedge v_k := ... • 7,531 3 votes Accepted ### Operator inner product This is not true. Indeed, here is a counterexample:$$A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \; v = e_1 \otimes e_2, \; \langle v, A \otimes Av \rangle = -1$$Here is a suggested ... • 9,745 3 votes Accepted ### What is the relation between the tensor of two R-modules over different rings. A sufficient condition is that the map \mathbb{Z} \to R is an epimorphism; this includes the case of quotients but is more general since it also includes localizations and some other stranger things.... • 431k 3 votes Accepted ### Definition of tensor product seems to contradict universal property Tensor products A canonical definition of a tensor product is by the universal property: The tensor product of the vector spaces U, V is a vector space U \otimes V, together with a bilinear ... • 11.1k 2 votes Accepted ### Why is the set of decomposable tensors closed under any norm-induced topology? On finite-dimensional spaces, all norms are equivalent, so we might as well pretend V_i are all inner product spaces and the norm on V_1 \otimes \cdots \otimes V_k is induced by the usual inner ... • 9,745 2 votes Accepted ### For two irreducible modules V and W, f : V\to W is a G-module isomorphism \iff \text{span}(f)\subset V^*\otimes W is trivial Perhaps it is better to identify V^* \otimes W with Hom(V,W). The action of G on Hom(V,W) is then seen to be$$ g \cdot f := (g \cdot f)(v) = g \cdot f(g^{-1} \cdot v)$$Recall now the ... • 401 1 vote ### Grounding the concept of a Free Vector space of the cartesian product of two vector spaces Your comment is correct, you can start with a 2-element basis \{v_1,v_2\} for V and a 3-element basis \{w_1,w_2,w_3\} for W and from them you can produce a 6 element basis for V \otimes ... • 124k 1 vote ### Two definitions of antisymmetrization of a tensor? Rewritten shorter but more rigorous answer: You have to distinguish between a vector space V and its dual V^*, which is a derived object. V^*\otimes V^* is easier to define than V\otimes V. It ... • 8,502 1 vote Accepted ### Constructing a basis for a tensor product If the pair (V\otimes W,\otimes) is a tensor product of vector spaces V and W with tensor product map \otimes:V\times W\to V\otimes W, then the pair satisfies the universal property: For any ... • 3,291 1 vote Accepted ### Do tensor products remain unchanged along ring epimorphisms? Yes. The easiest way to see this is to write everything in terms of tensor products over S, which means M \otimes_R N is rewritten$$M \otimes_S \left( S \otimes_R S \right) \otimes_S N$$and we ... • 431k 1 vote Accepted ### Double Trace of the tensor product of the metric tensor with vector fields. The notation you are using partly is not really well defined. There are (at least) two possible definitions of what tr[g\otimes X\otimes Y] should mean, namely either g(X,\ )\otimes Y or g(\ ,Y)\... • 21.2k 1 vote ### Discrete spectrum of A \otimes 1+ 1 \otimes B Let's complete the hints of Christian Remling to an actual answer. As shown here, the spectral measure E of A\otimes 1+1\otimes B satisfies$$ E(S)=\int_{\sigma(B)}\int_{\sigma(A)}1_S(\lambda+\mu)\...
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