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4

The fact that $\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{R}\cong\mathbb{C}$ only as $\mathbb{Z}$-modules follows as $\mathbb{Z}[i]\cong\mathbb{Z}^{\oplus 2}$ and $\mathbb{Z}^{\oplus 2}\otimes_{\mathbb{Z}}\mathbb{R}\cong\mathbb{R}^{\oplus 2}\cong\mathbb{C}$. If you need to use the universal property of tensor products somewhere, it can be used in showing that ...


4

The map you wrote is not an isomorphism. It is instead a map that arguably parametrizes (though not uniquely) all multiples of the diagonal map $M\rightarrow M^{\oplus n}$ (where the multiple, depending on an element $(r_1,\ldots,r_n)$ of $R^{\oplus n}$, is $\prod_ir_i$). This is why, as you observe, you could make a similar argument for $R^{\oplus n}\times ...


3

The set of $R$-bilinear maps $f: R\times N\to P$ is in natural bijection with the set of $R$-linear maps $g:N\to P$. The correspondence is obtained by setting $f(r,n) = g(rn)$.


3

Suppose $\alpha_{m+1}f(v_{m+1})+\dots+\alpha_nf(v_n)=0$. This can be rewritten as $$ f(\alpha_{m+1}v_{m+1}+\dots+\alpha_nv_n)=0 $$ that is, $$ \alpha_{m+1}v_{m+1}+\dots+\alpha_nv_n\in\ker(f) $$ In particular, $$ \alpha_{m+1}v_{m+1}+\dots+\alpha_nv_n= \beta_{1}v_1+\dots+\beta_mv_m $$ for some scalars $\beta_1,\dots,\beta_m$. Therefore ...


3

The definition of the tensor product is $(f \otimes g)(u,v) = f(u)g(v)$. You should be able to check that whenever $f,g \in T^*$ (i.e. are linear), their product $f \otimes g$ is bilinear; and the product $f \otimes f$ is symmetric and weakly positive definite. Thus the action of $dx \otimes dx$ on a pair $u,v$ of tangent vectors is $$(dx \otimes dx)(u,v) = ...


3

One reason why one needs the bilinear map (or multilinear in general), instead of just going ahead to define a map on the tensor product, is that one needs to show that the latter is well-defined. So it may be easy to define some $A \otimes B \rightarrow C$, but it actually may be quite hard to prove that it is well-defined. On the other hand, once a ...


2

The proof constructs two subalgebras $C$ and $D$ of $A$. You want to show that $C\otimes D\simeq A$, so you need to give an isomorphism $C\otimes D\to A$. The tensor product of two algebras $X$ and $Y$ has the following universal property : for any algebra $Z$, any two morphisms $f: X\to Z$ and $g: Y\to Z$ such that $f(x)g(y)=g(y)f(x)$ for all $x\in X$, ...


2

Consider $G=\Bbb Z[x], H= \Bbb R$ treated as $\Bbb Z$-modules. Then for everything to be an elementary tensor would mean that every polynomial in $\Bbb R[x]\cong\Bbb Z[x]\otimes_{\Bbb Z}\Bbb R$ is of the form $r\cdot p(x)$ for some $r\in\Bbb R$ and $p(x)\in\Bbb Z[x]$.


2

In general this is not true: for instance that $\mathbb{Z}_p \otimes_\mathbb{Z} \mathbb{Z}_q \cong \{ 0\}$ if $\gcd(p,q)=1$.


2

No. To show it's injective, you have to show that, if $\;\sum r_i\otimes m_i\mapsto\sum r_im_i=0$, then $\;\sum r_i\otimes m_i=0$. But that is because $\;\sum r_i\otimes m_i=\sum 1\otimes r_im_i=1\otimes0=0$.


2

An element of the domain of $T$ is of the form $(f_1,\dots, f_k, v_1, \dots, v_l),\, f_1,\dots,f_k\in V^*, v_1,\dots,v_l \in V$, so the first $k$ elements of $T$ should be able to take elements of $V^*$ and the next $l$ terms should be able to take elements of $V$, hence $v_{\mu_1} \otimes \cdots \otimes v_{\mu_k}(f_1,\dots, f_k)$ first, and $v^{{\nu_1}^*} ...


2

The key is that, for any injective representations (i.e., $*$-homormophisms, so in particular $f$ and $g$), $$ \|\sum a_j\otimes b_j\|_\min=\|(f\otimes g)\left(\sum a_j\otimes b_j\right)\| $$ (technically, this might require using a further set of faithful representations to get embeddings $A\hookrightarrow B(H)$, $B\hookrightarrow B(K)$, but it doesn't ...


2

HINT: show that every element of $\mathbb{Z}/m\mathbb{Z}\otimes\mathbb{Z}$ is equal to one of the form $[k]\otimes 1$ for some $k\in\{0, . . . , m-1\}$. It will be enough to show this for elements of the form $[x]\otimes y$.


2

Here's a counterexample. Take $A=k[x]/(x^2)$ and $M=N=A/(x)$. Then $\operatorname{Hom}(M,A)\cong M$, generated by the map $f:1\mapsto x$, and $\operatorname{Hom}(A,M)\cong M$, generated by the map $g:1\mapsto 1$. The tensor product $\operatorname{Hom}(M,A)\otimes \operatorname{Hom}(A,M)$ is then also isomorphic to $M$, generated by $f\otimes g$. But ...


2

Let $U$ and $V$ be free modules over a nontrivial commutative ring $R$ and let $\alpha:U\to R$ and $\beta:V\to R$ be two $R$-linear maps which have $1$ in their image; such things are easily seen to exist using freeness. Then using the properties of tensor products you can show that there is a morphism of abeelian groups $f:U\otimes_RV\to R$ such that for ...


1

The notation in the C$^*$ literature for tensor products is far from uniform. But yes, as $K(H)$ is nuclear, it doesn't matter which C$^*$-norm you put on $K(H)\otimes A$, they are all the same.


1

It's much easier to show that if $A$, $B$ and $C$ are $R$-modules (commutative $R$, for simplicity), then $$ (A\oplus B)\otimes_R C\cong (A\otimes_R C)\oplus (B\otimes_R C) $$ The bilinear map $(A\oplus B)\times C$ is $$ ((a,b),c)\mapsto (a\otimes c,b\otimes c) $$ and it's quite easy to show that this satisfies the universal property. By induction, we get ...


1

Intuitively, any $R$ bilinear map $f:R \times N \to M$ is really a linear map from $N$ to $M$ in disguise, for if $g$ is a linear map from $N$ to $M$, then we may define $f$ by letting $f(1,n) = g(n)$, extended to a unique bilinear extension. The details are then trivial. Let $f: R \times N \to M$ be a bilinear map. Given your projection map $(r,n) \mapsto ...


1

how exactly to represent specific linear transformations in tensor notation. Indeed every linear map $A:V\rightarrow W$ has a representation as a tensor $T\in V^{*}\otimes W$ such that $\forall x\in V:Ax=T(x)\in W$. In chosen bases, $$ T=\sum a_{ij}\;v_j^*\otimes w_{i} $$ where $a_{ij}$ is the matrix of the linear map and $\{v_j^*\}$ is the basis in ...


1

We have that $\mathbb Z[X]\otimes_\mathbb Z\mathbb R\cong \mathbb R[X]$ because both are free commutative $\mathbb R$-algebras over one element. It immediately follows that $\mathbb Z[i]\otimes_\mathbb Z\mathbb R\cong \mathbb R[i]\cong\mathbb C$


1

Here is another way to do it: you can just write down a module homomorphism $\mathbb{Z}[i]\otimes \mathbb{R}\to \mathbb{C}$ by $a+bi\otimes r \mapsto ra+rbi$, and show that it is an isomorphism. To see injectivity for example, you can say that if $ra + rbi = 0$ then $ra = rb = 0$ and hence $r = 0$ or $a=b=0$ in which case the tensor $a+bi\otimes r = 0$, so ...


1

Ok, lets formalise all that have been said in the comments: Let $(v_1,\dots,v_n)$ and $(v^1,\dots,v^n)$ be basis of $V_p$ and $V^*_p$, respectively. Take a tensor $\tau \in \mathcal{T}^r_s(V_p)$, and pick indexes $k \leq r, l \leq s$. Then we define the contraction $C^k_l\tau \in \mathcal{T}^{r-1}_{s-1}(V_p)$ as ...


1

Tensor products of c.p. faithful maps with respect to the minimal norm are faithful: D. Avitzour, Free products of C*-algebras, Trans. Amer. Math. Soc. 271 (1982), 423–435. Certainly, *-homomorphisms are c.p.


1

For finite dimensional vector spaces, we have an isomorphism $V \cong V^{**}$, so you are really using a basis of $V^{**}$


1

So with the help of @Winther, the solution is as follows: $$ U = \frac{1}{r^3}\left(\vec{P}\cdot\vec{Q} - \dfrac{3(\vec{P}\cdot\vec{r})(\vec{Q}\cdot\vec{r})}{r^2}\right) \equiv \frac{1}{r^3}P_iT_j^iQ^j\\ \vec{P}\cdot\vec{Q} - \dfrac{3(\vec{P}\cdot\vec{r})(\vec{Q}\cdot\vec{r})}{r^2} = P_iT_j^iQ^j\\ $$ $$ P^ke_kQ_le^l- ...


1

Not necessarily. A counter-example would be the sub-algebra of the algebra of $3\times 3$ matrices described by $$ A = \begin{pmatrix} k & k & k \\ 0 & 0 & k \\ 0 & 0 & k \end{pmatrix}. $$ For this algebra, $\mu:A\otimes A\to A$ is surjective, but there cannot be an identity element, since there are no $X\in A$ such that $$ X\cdot ...


1

Here is a very low dimensional example: consider a two dimensional vector space $V$ with basis $\left\{v_1,v_2\right\}$. Then $\left\{v_1\otimes v_1, v_2\otimes v_1,v_1\otimes v_2, v_2\otimes v_2\right\}$ is a basis of $V\otimes V$. You can easily show that $$v_1\otimes v_2+v_2\otimes v_1\neq u\otimes w$$ for all $u,w\in V$. Edit: Be sure to work out the ...


1

Tensor product of vector spaces and algebras are both special cases of tensor product of modules. Indeed a vector space is just a module over some field and an algebra is just a module that has a ring structure. So you really need to understand the tensor product of modules to get the big picture. To make things easy it is better to focus on commutative ...


1

Tensor product is an "external" operation. It takes two completely unrelated algebras $A$ and $B$ and it spits out a new algebra $A \otimes B$ that only depends on the data of $A$ and $B$ taken separately. $A$ and $B$ could be the same algebra, subalgebras of a bigger algebra, it won't matter from the point of view of $\otimes$. The product you define is an ...


1

(This works only for commutative $k$-algebras.) Hint. One can define a ring homomorphism $A\otimes_kA\to A$ by $\sum a_i\otimes b_i\mapsto\sum a_ib_i$.



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