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7

No, your argument isn't correct: if $1\otimes x=1\otimes y$ then not necessarily $x=y$. I'd do this as follows: if $M$ is a $\mathbb Z$-module, then $\mathbb Q\otimes_{\mathbb Z}M\simeq S^{-1}M$, where $S=\mathbb Z-\{0\}$. The isomorphism is given by $\dfrac ab\otimes x\mapsto\dfrac{ax}{b}$. This way $1\otimes(1,1,\dots)$ corresponds to the fraction ...


2

When you talk about the eigenvalues of $A^{(2)}$ over $V^{(2)}$, you assume that the image of $A^{(2)}$ lies within $V^{(2)}$ (or, in other words, that $V^{(2)}$ is an invariant subspace of $A \otimes A$). Note, however, that this is not generally the case. For example, we can consider the map $A$ over $\langle x_1,x_2 \rangle$ defined by $$ Ax_1 = Ax_2 = ...


2

$V$ is a $k[T]$-module. $V \otimes $ Something is a ($k[T] \otimes$ Something) - module.


2

Actually, matrix multiplication can be seen as a bilinear map $$K^{n\times k}\times K^{k\times m}\rightarrow K^{n\times m},$$ where $K$ is the ground field. Such a bilinear map can be expressed as a tensor of order 3 of format $(nk)\times (km)\times (nm)$ (any bilinear map $K^{\ell_1}\times K^{\ell_2}\rightarrow K^{\ell_3}$ can be represented by a tensor of ...


2

Since $N$ is finitely generated, we have an exact sequence $$M\stackrel{f}{\rightarrow} N\rightarrow \mathbb{Z}^k\times T\rightarrow 1$$ where $T$ is finite and $k \ge 0$. If $f$ is not surjective then either $k\ge 1$ or $T$ is nontrivial. Since tensoring is right-exact, tensoring this sequence with $\mathbb{Z}/p$ gives: ...


2

The algebra of polynomials on a vector space $W$ (of dimension $N$, say) is just a coordinate-free way of saying "polynomials in $N$ variables". So, if you like, a polynomial on $W$ is just a function $q:W\to \mathbb{R}$ such that if you pick a linear isomorphism $f:\mathbb{R}^N\to W$, the composition $qf:\mathbb{R}^N\to\mathbb{R}$ is a polynomial function ...


2

For any $k[\mathbb{S}_n]$-module $M$, $M\otimes_{k[\mathbb{S}_n]}k$ is the largest quotient of $M$ on which $\mathbb{S_n}$ acts trivially. Factoring $V^{\otimes n}$ by the ideal generated by differences of elements that differ just by swapping two tensor factors gives the largest quotient on which transpositions act trivially. So it comes down to the fact ...


2

The answer is negative. Let $C=K[X^2,X^3]$, and $A=B=K[X]$. Then $A'\otimes_{C'}B'=K(X)\otimes_{K(X)}K(X)=K(X)$ is a field, while $A\otimes_CB$ is not an integral domain. In order to show the last claim notice that $$[X\otimes(X+1)-1\otimes X(X+1)][X\otimes(X+1)+1\otimes X(X+1)]=0$$ and use this answer where I've proved that the factors are both non-zero. ...


2

Note that the trace of any permutation is its number of fixed points (that is, the number of $1$-cycles). As in my previous answer, let $x_{ij}$ denote $x_ix_j = x_{ji}$. In order to have $A^{(2)}(x_{ij}) = x_{ij}$, we must have $$ (Ax_i)(Ax_j) = x_{ij} \DeclareMathOperator{\tr}{Tr} $$ There are precisely two ways in which this can occur: $x_i$ and $x_j$ ...


2

I'll write the two coordinate systems as $x_j$ and $y_k$, to better distinguish them. Then we know that $a_j \frac{\partial}{\partial x_j}=b_k \frac{\partial}{\partial y_k}$. So, by applying both sides to the coordinate function $x_i$, we know that $a_j \frac{\partial x_i}{\partial x_j}=b_k \frac{\partial x_i}{\partial y_k}$. By definition, $\frac{\partial ...


2

This is false, even when $G$ is finite but it's not far off from being true. If $G$ and $H$ are finite groups, irreducible representations of $G \times H$ are of the form $\{V \otimes W\}$ with $V, W$ irreducible representations of $G, H$ respectively. So, representations of $G \times H$ are direct sums of tensor products of representations of $G$ and $H$. ...


1

Your "ternary dot product" is basically equivalent to the Hadamard product (https://en.wikipedia.org/wiki/Hadamard_product_%28matrices%29): given two vectors $\langle a_i\rangle, \langle b_i\rangle$, their Hadamard product is just componentwise multiplication: $$\langle a_i\rangle*\langle b_i\rangle=\langle a_ib_i\rangle.$$ To see why this is equivalent to ...


1

You could call it, for example, ${\bf x}^T Y {\bf z}$ where $Y$ is the diagonal matrix with the entries of ${\bf y}$ on the diagonal. Of course you could apply any permutation to ${\bf x}$, ${\bf y}$, ${\bf z}$.


1

Yes, such a kind of exact decomposition is always possible. This is shown in Barenco et al., Elementary gates for quantum computation (Sec. 8, pg. 27 in the arXiv preprint), based on work by Reck et al. (which gives a corresponding decomposition where each elementary operation is the identity except for a $2\times 2$ submatrix). The construction given ...


1

If $A$ acts on this basis than the trace is simply going to be the size $N$ of the set of basis vectors that it fixes (i.e. the number of $i$ such that $A(x_i) = x_i$) and as such we have that $Tr(A) = \sum_{i = 1}^n \lambda_i = N$. Then the number of basis vectors fixed by $A^{(2)}$ is simply $\frac{\text{Tr}(A^2) + \text{Tr}(A)^2}{2} - N$ which is ...


1

There are often different ways of viewing the same sort of tensor, and people will often not bother to make the distinction. Natural isomorphisms between the operations that proudce the various tensors spaces justify this identification. In fact some notations for working with tensors will not even make a choice as to which viewpoint is being taken. Here is ...


1

Consider the map $g: R/(I+J)\to R/I \otimes_R R/J$ given by $\overline{a} \mapsto \overline{a} \otimes 1$. If you confirm for yourself that this is a well-defined function, it's obvious that it's an inverse to your map $f$.


1

We could write your tensor as $$ (a_1,a_2) \otimes (b_1,b_2) \otimes (c_1, c_2) $$ Or, depending on your notation, perhaps $$ (b_1,b_2) \otimes (c_1, c_2) \otimes (a_1,a_2) $$ One way to check that this tensor is rank one is to note that one matrix is a multiple of the other, and that each matrix is rank one.


1

Let $R$ be a commutative unitary ring, $M$ an $R$-module, $N$ a finitely generated $R$-module, and $f:M\to N$ a homomorphism. Suppose that $\bar f:M/\mathfrak mM\to N/\mathfrak mN$ is surjective for every maximal ideal $\mathfrak m$ of $R$. Then $f$ is surjective. $f$ surjective $\Leftrightarrow$ $f_{\mathfrak m}:M_{\mathfrak m}\to N_{\mathfrak m}$ ...


1

No, that is another pair of shoes. It has nothing to do with the fact that von Neumann algebras are dual spaces. Short answer: $M\odot N$ denotes the algebraic tensor product of $M$ and $ N$, which has the natural structure of a *-algebra. You probably know that each von Neumann algebra has a normal, faithful representation on some Hilbert space. Let ...



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