# Tag Info

6

If you want an honest functor, you can simply choose any representative of the isomorphism class in question; it doesn't have to be "canonical". That is, for every pair $(X,Y)$ of $Z$-schemes, you can choose some triple $(P_{X,Y},p_{X,Y},q_{X,Y})$ where $P_{X,Y}$ is a $Z$-scheme and $p_{X,Y}:P_{X,Y}\to X$ and $q_{X,Y}:P_{X,Y}\to Y$ are maps of $Z$-schemes ...

4

The tensor product is defined up to canonical isomorphism, which is different from being defined up to isomorphism: there is a universal map $M\times N \to M\otimes N$ through which any bilinear map $M \times N \to R$ factors, and if you give one construction of the tensor product (e.g. generators and relations), and I give another one (e.g.perhaps I ...

4

$\mathbb R$ is a faithfully flat $\mathbb Q$-module (since it is free). Then the sequence $0\to A\otimes_{\mathbb Z}\mathbb Q\to B\otimes_{\mathbb Z}\mathbb Q\to0$ tensorized by the $\mathbb Q$-module $\mathbb R$ is exact, so it is exact.

3

Yes, this is true. Let $S$ be a basis for $A$; then for any group $G$, there is a natural isomorphism $A\otimes G\cong G^{\oplus S}$, where $G^{\oplus S}$ denotes the direct sum of copies of $G$ indexed by $S$, i.e. the group of all functions $S\to A$ which are $0$ at all but finitely many points (this is because $A\cong\mathbb{Z}^{\oplus S}$, ...

3

Let $D$ be a tensor derivation on a manifold $M$ I assume it means $D$ is $\mathbb{R}$-linear and obeys Leibniz rules with respect to tensor multiplication and contraction See axioms 2 and 3 here. Actually here we only need 2 things: (1) For every vector field $X$ and 1-form $\omega$ we have $$D(\omega(X))=D(\omega)(X)+\omega(D(X)).$$ (2) For function ...

3

They're the same. The fact you want to use to show this is that if $f$ is a bilinear map from $V \times V$ to $R$, then it extends to a map on your group $G$, and vanishes on all of the elements of $H$, so it induces a linear map on the tensor product $V \otimes V$. Knowing this, we want the linear maps $L(V \otimes V, R)$ to correspond to the bilinear maps ...

3

Not sure if that is what you want. There is an obvious map $$\Gamma(E_1) \otimes_\mathbb{R} \Gamma(E_2) \to \Gamma(E_1) \otimes_{C^\infty(M)} \Gamma(E_2)$$ which sends $s_1 \otimes _{\mathbb R} s_2$ to $s_1 \otimes_\infty s_2$ (write $\otimes_\infty = \otimes_{C^\infty(M)}$ for simplicity). This map is never an isomorphism (as $\mathbb R$-module), ...

2

By definition, we have that $$R^{mn}\cong Re_{11}\oplus...\oplus Re_{1n} \oplus Re_{21}\oplus ... \oplus Re_{2n} \oplus ... \oplus Re_{m1} \oplus ... \oplus Re_{mn}$$ That is, $R^{mn}$ has a basis consisting on $mn$ elements which we have chosen to denote by $\{e_{ij}\}$, where $i=1,...,m$ and $j=1,...,n$. In the same fashion, pick $\{a_1,...,a_m\}$ a ...

2

The universal property of $R^J$ for a set $J$ is that there is a natural isomorphism $$\hom_R(R^J,M)\simeq \hom_{\rm Set}(J,M)$$ Now, by the usual adjunctions in ${}_R\,\mathbb {mod}$ and $\mathrm{Set}$ we have natural isomorphisms $$\hom_R(R^I\otimes_R R^J,M)\simeq \hom_R(R^J,\hom_R(R^I,M))\\\simeq \hom_{\rm Set}(J,\hom_{\rm Set}(I,M))\simeq \hom_{\rm ... 2 I think it's much more common to let the last index position be the one introduced by differentiation, not the first. But regardless of which convention you choose, you're never going to have a Leibniz rule for total covariant derivatives of the form \nabla (T\otimes S) = \nabla T \otimes S + T\otimes \nabla S. (Notice, for example, that if both S and ... 2$$M\otimes_RN=0\Rightarrow M\otimes_RN\otimes_RR/m=0\Rightarrow M\otimes_RN/mN=0\Rightarrow (M\otimes_RR/m)\otimes_{R/m}N/mN=0\Rightarrow M/mM\otimes_{R/m}N/mN=0$$2 You mean r\otimes(m+ni)\mapsto r(m+ni). The inverse map is a+bi\mapsto a\otimes1 + b\otimes i. 2 If a group acts on a set X and Y is a subset of X then I would nornally interpret the "subgroup of G that fixes Y" to mean \{ g \in G : g(y) \in Y\,\forall y \in Y \}. There is also the possibility that it could mean \{ g \in G : g(y) = y\,\forall y \in Y \}, but my guess is that this is not what is meant here. In your example, X is the ... 1 Do you know anything about localizations? \mathbb{Z}_{(p)} is the localization of \mathbb{Z} at the multiplicatively closed subset S=\mathbb{Z}\setminus p\mathbb{Z}. Now you can use the fact that localization commutes with tensor products,i.e, S^{-1}\mathbb{Z} \bigotimes A = \mathbb{Z} \bigotimes S^{-1}A = S^{-1}A = A, as A is an abelian-p ... 1 If M is flat, then it must be torsion-free (see here), but this is not the case: xy=0 and y\ne 0 in M. 1 The first attempt is fine. The inverse you describe is well defined and the maps are in fact mutually inverse. Let's have a look at the second approach. (Edited for clarification.) Tensoring the exact sequence$$0\to \mathfrak{a}\to A\to A/\mathfrak{a}\to 0$$with M yields the exact sequence$$\mathfrak{a}\otimes_A M\to A\otimes_A M\to ...

1

I basically had the same question as you. When defining the covariant derivative as a derivation, then $\nabla (S \otimes T)$ doesn't seem to make sense in some case unless one rearranges the covariant and contravariant factors in S and T, and I don't know if there is a canonical way to do this... Even in certain cases where no such rearrangement is ...

1

Yes, your understanding is correct. In general, if $f: A\to B$ and $g: C\to D$ are morphisms then the map $f\otimes g: A\otimes C\to B\otimes D$ is defined to be the unique morphism satisfying $(f\otimes g)(a\otimes b) = f(a)\otimes g(b)$ (uniqueness then follows from the universal property of the tensor product).

1

I'm not sure what you count as "nice", but here's a counterexample. Let $k=\mathbb{R}$ and $A=B=M=N=\mathbb{C}$. Then $A\otimes_k B=\mathbb{C}\otimes_\mathbb{R}\mathbb{C}$ is not simple as a module over itself, being the direct sum of the submodules generated by $i\otimes 1-1\otimes i$ and $i\otimes 1+1\otimes i$. More generally, if $A=B$ is a field ...

1

These definitions coincide if $V$ is a finite dimensional vector space. If we denote by $V\otimes V$ the tensor product via the quotient construction, then you can construct an isomorphism $\Phi\colon V\otimes V\rightarrow L(V^*,V^*,\mathbb R)$ as follows: The natural map $V\times V\rightarrow L(V^*,V^*,\mathbb R),\; (v,w) \mapsto [(\varphi,\psi)\mapsto ... 1 By definition, the rank of a tensor$t$is given by is defined to be the smallest$k$such that we can find$v_1,\dots,v_k \in V$and$w_1,\dots,w_k \in W$satisfying $$t = \sum_{j=1}^k w_j \otimes v_j$$ The key insight to this problem, however, is that the rank of a map can be defined similarly. The rank of a map$\phi \in \operatorname{Hom}(V,W)\$ is ...

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