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4

Yes, of course. After all universal objects are unique up to canonical isomoprphism so you should be able to do all (or at least most) abstractly. For example $V\otimes W\cong W\otimes V$ follows immediatly from $V\times W\cong W\times V$ via $(v,w)\mapsto(w,v)$, which respects bilinearity. That the $v\otimes w$ span $V\otimes W$ follows from the fact that ...


3

The general situation is the following: $A$ is a commutative ring and $V_1, V_2, W_1, W_2$ are $A$-modules If one of the ordered pairs $(V_1,V_2)$, $\,(V_1,W_1\,$ or $\,(V_2,W_2)$ consists of finitely generated projective $A$-modules, the canonical map is an isomorphism (Bourbaki, Algebra, Ch. 2 ‘Linear Algebra’, §4, n°4, prop. 4). Over a field, all ...


3

I don't think what you say is true. $\mathcal{L}(V) = \text{Hom}(V, V) = V \otimes V^*$. Your map $\Theta_v: V \otimes V^* \rightarrow V$ is given by tensoring the evaluation map $ev_v: V^* \rightarrow k$ with $V$. A functional on $\text{Hom}(V, V)$ is the same as an element of $\text{Hom}(V, V)$, the relationship being taking the dual of an endomorphism ...


3

A different reason why it can't be true is that there just aren't enough $(\varepsilon,v)$ pairs to represent all the functionals when $F$ is finite. Suppose for example $F=\mathbb F_p$ and $n=3$. $\mathcal L(V)$ is a 9-dimensional vector space and therefore there are $p^9$ different linear functionals on $\mathcal L(V)$. On the other hand there are only ...


3

In fact $a\otimes b\mapsto a\otimes b$ is an isomorphism $A\otimes_DB\leftrightarrow A\otimes_FB$ for any spaces (or algebras) $A$ and $B$ over a field $F$ which is the fraction field of a domain $D$. The reason is that $$\begin{array}{ll} \displaystyle \color{Blue}{\frac{1}{y}}a\otimes b & \displaystyle =\frac{1}{y}a\otimes \color{Green}{y}\frac{1}{y}b ...


3

$g$ is not assumed to be linear. The purpose is to eventually show that $g:M\times N\to P$ is the universal bilinear map characterizing $P$ as the tensor product of $M$ and $N$.


2

Consider the following commutative diagram of $\Omega$-modules with exact rows. $$\require{AMScd} \begin{CD} 0 @>>> I_k \otimes_k \Omega @>>> k[x_1,\ldots,x_n] \otimes_k \Omega @>>> (k[x_1,\ldots,x_n]/I_k) \otimes_k \Omega @>>> 0 \\ \ @VVV @VV\sim V @VVV \ \\ 0 @>>> I @>>> \Omega[x_1,\ldots,x_n] ...


2

One reason is this leads to modules that behave poorly with respect to commutators. For example, let $m\in M$, $n\in N$, and suppose $r,s\in R$ don't commute. Then $$rm\otimes sn=rs(m \otimes n)=sr(m \otimes n)$$ So any commutator $rs-sr$ annihilates the whole tensor product.


2

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 ...


2

Because $\text{Sym}_2(\Bbb C^2) \cong \Bbb C^3$: The vector space of symmetric $2\times 2$ matrices is $3$-dimensional.


2

Hint: for the first one, tensor product is distributive with respect to a finite sum. As for the second, write $1=\frac{n}{n}$ in $\mathbb{Q}$.


1

Short answer: yes. In what sense do you mean that the tensor product is commutative? That switching $V$ and $W$ in $V\otimes W$ gives a space isomorphic to $W\otimes V$? Then we can show this in the following way: $W\otimes V$ satisfies the same universal property as $V\otimes W$ so they must be isomorphic. Consider any bilinear map $f$ from $V\times W \to ...


1

To show things like this in the case of finite dimensional vector spaces, you can always fallback on picking bases: Let $v_i^1, \ldots, v_i^{m_i}$ be a basis of $V_i$ ($i=1,2$), and $w_i^1, \ldots, w_i^{n_i}$ be a basis of $W_i$ ($i=1,2$). Let $f^{jk}_i : V_i \to W_i$ be given on the basis by $f_i^{jk}(v_i^l) = \delta_{jl} w_i^k$. Then $\{v_1^p \otimes ...


1

$(|1\rangle \otimes |0\rangle \otimes |1\rangle) + (|0\rangle \otimes |0\rangle \otimes |1\rangle)$ is clearly a product state. It's tricky to find good notation to denote this and I don't think there's an established standard way. Here's one option: Let $\sigma_{ij}$ be an automorphism that flips the $i$th and $j$th position of a tensor product ...


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 ...


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

Suppose $G$ acts on a vector space $V$ with character $\phi_1$, and $H$ acts on a vector space $W$ with character $\phi_2$. Then $G \times H$ acts on the vector space $V \otimes W$ with character $\Phi(g,h) = \phi_1(g)\phi_2(h)$. Moreover every irreducible complex representation of $G \times H$ is a tensor product of an irreducible representation of $G$ ...


1

Atiyah-Macdonald has: In this case $M\otimes_R R_f\cong M_f$, (for which $m\otimes 1/1\mapsto m/1$). Now, because of $m\otimes 1/1=0,$ (in $M\otimes_R R_f$), you have $m/1=0$ in $M_f$. So there exists $f^n$ such that $f^n m=0$ in $M$.



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