# Tag Info

9

The grammar is "a category is tensored over a monoidal category"; this is a generalization of a set being equipped with an action of a monoid, or an abelian group being equipped with an action of a ring. In full generality you should provide the tensoring, but sometimes if you require enough it exists uniquely. The general pattern of the uniqueness results ...

4

$v\otimes w$ is either of the following: The rank 1 linear mapping $V^*\to V$ given by $\alpha\mapsto v.\langle \alpha,w\rangle$ where $\langle\;,\;\rangle:V^*\times V\to \mathbb F$ is the duality. This is the easiest to visualize. The decomposable bilinear form $V^*\times V^*\to \mathbb F$ given by $(\alpha,\beta)\mapsto \langle\alpha,v\rangle ... 4 Intuitively, if a functor 'forgets structure' from one category of sets-with-algebraic-structure to another then it will have a left-adjoint; such adjunctions are called 'free-forgetful' adjunctions. Unfortunately$F$is an unfortunate choice of letter because forgetful functors are usually denoted by$U$and free functors by$F$, so for your$F$I'll write ... 3 The point is that in contrast to a short exact sequence, a split short exact sequence can be viewed as a certain kind of diagram with additive commutativity relations: Definition. A sequence$A\xrightarrow{i} B\xrightarrow{\pi} C$is split short exact if there exist$B\xrightarrow{r} A$,$C\xrightarrow{\sigma} B$such that $$ri=\text{id}_A,\quad ... 2 The internal hom is defined via maps into it. You know nothing about the classification of maps into a coproduct, but you know how to classify the maps into a product, by the very definition of a product. You can also see how the product comes up in the following calculation: Let A,B,C be graded R-modules (the case of chain complexes is similar, but ... 2 Of course, the algebraic intuition should be the one given by Clive Newstead. However, it might be worth noting that your problem is in fact an instance of something more general. Take \mathcal C to be a category with pushouts. Denotes {}_{x \backslash}\!\mathcal C the category whose objects are the morphisms x \to a of \mathcal C and whose ... 2 Let B be an A-algebra and M, N be B-modules. Here is an example in which M \otimes_A N and N \otimes_A M are isomorphic as A-modules, but not as B-modules. \mathbb Z[x] is a \mathbb Z-module via the inclusion \mathbb Z \hookrightarrow \mathbb Z[x]. Let M = \mathbb Z[x] be a \mathbb Z[x]-module via the identity. Let N = \mathbb ... 2 The ring B\otimes _A B has a canonical A-algebra structure and TWO different structures of B-algebra, which I'll call (B\otimes _A B)_l and (B\otimes _A B)_r, according as multiplication by elements of B happens on the left or on the right. These B-algebra structures are in general different since if we denote by \bullet and \circ the ... 2 In general, no. For instance, let A be a field, let$$M = N = \langle v_1, v_2, v_3, v_4 \rangle$$be a 4-dimensional vector space over A, and let$$ P = (M \otimes N) / \langle v_1 \otimes v_2 + v_3 \otimes v_4\rangle, $$with f the natural map. No non-zero decomposable tensor lies in the kernel$$\ker(f) = \langle v_1 \otimes v_2 + v_3 \otimes ... 2 Well, first you should recognize that we need some way of representing the$B$you have in your first definition in the second definition. This is because the first definition is the definition of a linear mapping with the thing it maps (a bivector) and the second definition is just a linear mapping without the thing that it maps. In fact the second ... 2 I think the key here is to understand what is meant by the tensor product. For any vectors$a, b, u, v$, the tensor product$a \otimes b$means $$(a \otimes b)(u,v) = (a \cdot u)(b \cdot v)$$ Or, perhaps, it might mean this instead: $$(a \otimes b)(v) = a (b \cdot v)$$ The two notions are equivalent to each other, so mathematicians freely use one or the ... 2 Let$\{0\}=N_0<N_1<\cdots<N_n=N$be a composition series for$N$. In particular,$N_i/N_{i-1}$is a cyclic module, so$N$can be generated by (at most)$n$elements. Then there is a surjection$R^n\to N$, and tensorizing by$M$we get$M\otimes_R R^n\to M\otimes_RN\to 0$. But$M\otimes_R R^n\simeq M^n$, so$M\otimes_R R^n$has finite length which is ... 2 It should be stated in the problem but I think you are right about$v_i = \pi (u_i)$. Hint: observe that$\frac{1}{2}((u_1 + u_2) \otimes (u_1 + u_2))= \frac{1}{2}(u_1 \otimes u_2 + u_2 \otimes u_1) + \frac{1}{2}(u_1 \otimes u_1 + u_2 \otimes u_2)$is projected by$\pi$to$0$on one hand (projection of LHS) and to$\frac{1}{2}(u_1 \otimes u_2 + u_2 \otimes ...

1

You are missing a term, we have$\def\tensor{\otimes}$ $$\nabla \cdot (u \tensor u) = (\nabla \cdot u)u + (u\cdot \nabla) u$$ By definition of $\nabla \cdot (u \tensor u)$, the $i$-th component of this vector is the divergence of the $i$-th row of $u \tensor u$. So \begin{align*} [\nabla \cdot (u \tensor u)]_i &= \nabla \cdot (uu_i)\\ ...

1

Note that if $g\circ f$ is the identity for some map $g: B\otimes_A B \to B$, then $f$ is injective for formal reasons. Namely, if $f(b_1) = f(b_2)$, then $b_1 = g(f(b_1)) = g(f(b_2)) = b_2$. The fact that $B$ is an $A$-algebra means that you have a good candidate for $g$.

1

Writing $G \cong \mathbb Z/n_1 \oplus \cdots \oplus \mathbb Z/n_r$, we can indeed reduce the problem to cyclic groups, since $$\text{Hom}_{\mathbb Z}(G,k^{\times}) \otimes_{\mathbb Z} \mathbb Z/p \cong \bigoplus_{i=1}^r \text{Hom}_{\mathbb Z}(\mathbb Z/n_i,k^{\times}) \otimes_{\mathbb Z} \mathbb Z/p,$$ which you can read either way by distributivity of the ...

1

I don't have a complete answer, but I think I can add some information. In the spirit of you last comment, the way toweards characterizing the cone of simple tensors is looking at maximal linear subspaces contained in it. Each line in $V$ determins a linear subspace of dimension $dim(W)$ contained in the cone and vice versa. These two families are exactly ...

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As in the question, we set $n := \dim V$ and $m := \dim W$ and assume both are finite (though many conclusions hold just as well when either or both are infinite), and pick arbitrary bases $A = (v_a)$ of $V$ and $B = (w_b)$ of $W$. Also, we usually regard bases as ordered sets of vectors; the below doesn't always specify orders, but of course these can be ...

1

Not a good place for a tensor product formalism since these Qx, Qy, Qz must be unbounded. Algebras of unbounded operators are not very useful, mathematically, due to restrictions related to domains of definition. As a general remark, we can use projector-valued measures from the spectral theory to formally define “Hermitian operators” with values in affine ...

1

It might be easier to approach this problem the following way: We have the following free resolution of $I$: $$0 \to R \to R \oplus R \to I \to 0,$$ where the first map is given by $1 \mapsto (-y,x)$ and the second map is given by $(1,0) \mapsto x, (0,1) \mapsto y$. After tensoring with $R/I$ we get the following exact sequence: 0 \to Tor_1(I,R/I) \to ...

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