# Tag Info

6

I will mention to two applications here and I hope you get some ideas and of course it may won't cover all cases. Let $A$ be set of alleles for the type $i$ of the gamete. You can look at our gamete as a string with $N$ types ( $L:=\{1,\cdots,N\}$ set of types) and every type can have an allele from $A_i$, set of possible alleles for type $i$. Assuming ...

4

$\mathbb Q \otimes_{\mathbb Z} \mathbb Q = \mathbb Q$ gives a negative answer to both questions.

3

${\bf NB:}$ This answer is for the original question before it was edited: how to change coordinates of the expression $\text{div}(A\nabla u)$ and put it in an explicit and compact form. Here is a very basic derivation of such a explicit and relatively compact form that you ask for which allows for a fairly simple implementation numerically. There do ...

3

It is an entagled state. We will prove it by contradiction. We will assume that your state is tensor product of 3 qubits. Take 3 qubits $a_1|0\rangle+b_1|1\rangle,a_2|0\rangle+b_2|1\rangle,a_3|0\rangle+b_3|1\rangle$ and multyply them tensornally. You get that $$(a_1|0\rangle+b_1|1\rangle)\otimes (a_2|0\rangle+b_2|1\rangle)\otimes ... 2 You can recover \phi from \nu'=(\mathrm{id}\otimes\phi)\circ \nu by applying \epsilon and using the relation between \epsilon and \nu. This forces \phi to be unique. In fact, you can just write down what \phi has to be:$$(\epsilon\otimes\mathrm{id})\circ(\mathrm{id}\otimes\nu')$$(Or rather \phi is the map X^*\rightarrow X^{\wedge} ... 2 How can you choose x_i^*,x_j^* among the elements of the dual basis? Well, they can either be different, which gives you \binom{n}{2} elements, or you can have x_i^*=x_j^*, giving you n more possibilities, so$$\dim\{x^*_i \otimes x^*_j + x^*_j \otimes x^*_i \mid x^*_i,x^*_j \in V^*\} = \binom{n}{2} + n.$$Now,$$\binom{n}{2} + \left(\binom{n}{2} + ...

2

Well, nothing special: write $C \in M_{mn}(\mathbb C)$. Let $C_{ab} \in M_{mn} (\mathbb C)$, where $a,b \in \{1, 2, \cdots, n\}$ so that $$(C_{ab})_{ij} = \begin{cases} C_{ij} & \text{if } (a-1)m +1 \le i\le am, (b-1)m+1\le j\le bm,\\ 0 & \text{otherwise.}\end{cases}$$ Abusing notations, we also consider $C_{ab} \in M_m(\mathbb C)$. Then $$C = ... 2 You can identify \mathcal T(\mathbb R) with the Algebra of Polynomials \mathbb R[x]. As stated in the answer of @Fallen_Apart, an element in  \mathcal T(\mathbb R) is just a finite sum of real number which have assigned a degree i (indicating in which of the spaces \mathcal T^i(\mathbb R)\cong\mathbb R they sit. Now you can just use these as the ... 2 The tensor product is \mathbb{F}_p[X, Y]/(r(X), s(Y)). More generally, tensor products of commutative rings can be computed by "concatenating" their presentations. This tensor product will usually fail to be a field. For example, \mathbb{F}_{p^n} \otimes \mathbb{F}_{p^n} turns out to be the direct product \prod_{i=1}^n \mathbb{F}_{p^n}. This is a ... 2 The expression S \otimes_A Hom_A(S,S) doesn't make sense, because if S is an irreducible right A-module, there is usually not any natural left A-module structure on Hom_A(S,S). For instance, if A=M_n(\mathbb{C}) for some n>1 and S=\mathbb{C}^n, then Hom_A(S,S)=\mathbb{C} cannot be made into an A-module (in any way compatible with the ... 2 Linear Independent Case A nice way of proving this is to use non-orthogonal basis for \mathbb{R}^3. Hence, consider the following definitions for the non-orthogonal basis$$\matrix{ {{{\bf{g}}_1} = A} & {{{\bf{g}}_2} = B} & {{{\bf{g}}_3} = C} \cr } \tag{1}$$and then the dual basis will be$$\matrix{ {V = \left( {{{\bf{g}}_1} \times ...

2

They are very useful in physics. For example, you need to use the tensor product to describe a system formed by two subsystems of spin $1/2$. You can find more information here. Also this Physics SE question: http://physics.stackexchange.com/questions/53039/when-and-how-did-the-idea-of-the-tensor-product-originate-in-the-history-quantum can be useful to see ...

2

A basic fact about category theory states that a morphism $f \colon A \to B$ is an epimorphism in a category $\mathcal C$ if and only if the pushout of it with itself (i.e., the cobase change of $f$ along itself) exists and $\require{AMScd}$ \begin{CD} A @>f>> B\\ @V f V V @VV \text{id}_B V\\ B @>>\text{id}_B> B \end{CD} is ...

1

$\DeclareMathOperator{\Sym}{Sym}$ $\DeclareMathOperator{\ker}{ker}$ $\DeclareMathOperator{\Im}{Im}$ I attempted to fix my voted up comment in another comment, but it got too long. It doesn't look like this question is getting any more attention, so let me put some true ( hopefully! ) things down. What I wrote in the comment is not really enough in the ...

1

Rewriting: $${\delta _{nm}}{\varepsilon _{ijk}} - {\delta _{im}}{\varepsilon _{njk}} - {\delta _{jm}}{\varepsilon _{ink}} = {\varepsilon _{ijn}}{\delta _{km}}$$ as: $${\delta _{nm}}{\varepsilon _{ijk}} = {\delta _{im}}{\varepsilon _{njk}} + {\delta _{jm}}{\varepsilon _{ink}} + {\delta _{km}}{\varepsilon _{ijn}}$$ makes it easier to see the symmetry. ...

1

Suppose that $p(x) = \sum a_i x^i$ and $q(x) = \sum b_j y^i$. Then we have $$p(x) \otimes q(y) = \sum_{i,j} a_i b_j (x^i \otimes y^j)$$ The tensor product of the modules $R[x] \otimes R[y]$ is the space of all polynomials of the form $$\sum_{i,j} c_{ij} (x^i \otimes y^j)$$ It may be helpful to think of this as the space of polynomials of the form $$... 1 If you feel uncomfortable with tensors, you can do everything in the setting of linear maps. Starting from a linear map A:V\to L(V,V), you obtain a skew symmetric, bilinear map \delta A:V\times V\to V via \delta A(v,w)=A(v)(w)-A(w)(v). Now if you have a linear subspace \mathfrak g\subset L(V,V), you can look at the restriction of \delta to ... 1 I represent my solution in some steps. As it may become too lengthy, I leave the proof of some parts to you. Also, in what follows, u is a scalar, {\bf{v}}, {\bf{a}}, {\bf{b}} are vectors, {\bf{A}}, {\bf{B}} are second order tensors (or Matrices). Furthermore, you should notice the following definitions$$\begin{array}{l} {\bf{A}}:{\bf{B}} = ...

1

Be careful. It's cleanest to describe the tensor-hom adjunction with three different rings instead of one, to make it as hard as possible to accidentally write down the wrong thing, so let $A, B, C$ be three different rings, let $_A M_B$ be an $(A, B)$-bimodule, let $_B N_C$ be a $(B, C)$-bimodule, and let $_A K_C$ be an $(A, C)$-bimodule. Then ...

1

There are better answers than mine since I have little experience with this. I'll just share how I approach these dimension questions. I think that finding a basis really is the way to go. The tensor algebra is graded by $T(V)=\bigoplus_{n\in\mathbb{N}} T_n(V)$, where $T_n(V)=V^{\otimes n}$ has a basis given by $n$-fold tensors of basis elements ...

1

A bit more detailed look at what Qiaochu said. Unfortunately my answer won't really be expressed in terms of $r$ and $s$. I hope it still helps you in some way. We know that $\Bbb{F}_p[X]\otimes \Bbb{F}_{p^n}\cong\Bbb{F}_{p^n}[X]$ and that $\Bbb{F}_{p^n}$ is a flat $\Bbb{F}_p$-module. Let us consider the short exact sequence ...

1

You said: Taking the outer product with the B vector: $\begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{bmatrix} \otimes \begin{bmatrix}1&0\end{bmatrix} = \begin{bmatrix}\frac{1}{\sqrt{2}}&0\\\frac{1}{\sqrt{2}}&0\end{bmatrix}$ But in your case $B=\begin{bmatrix}1\\ 0\end{bmatrix},$ so it should have been ...

1

To extend davcha's answer, for your specific example, you would get: $$A\otimes B= \left( \begin{array}{cc} \left( \begin{array}{ccc} 5 & 6 & 7 \\ 8 & 9 & 10 \\ \end{array} \right) & \left( \begin{array}{ccc} 10 & 12 & 14 \\ 16 & 18 & 20 \\ \end{array} \right) \\ \left( \begin{array}{ccc} 15 & 18 & 21 \\ ... 1 Your question is why$$\mathfrak m/\mathfrak m^2\otimes_R\mathfrak m/\mathfrak m^2\simeq \mathfrak m/\mathfrak m^2\otimes_{R/\mathfrak m} \mathfrak m/\mathfrak m^2. We have $\mathfrak m/\mathfrak m^2\simeq \mathfrak m\otimes_RR/\mathfrak m$ and $R/\mathfrak m\otimes_RR/\mathfrak m\simeq R/\mathfrak m$. Now all should be clear.

1

Yes, but not naturally. Both the invariants and coinvariants of a permutation module are free $R$-modules on the orbits of the action of $K$ on $X$. For the invariants, given an orbit $O$ the corresponding invariant is $\sum_{o \in O} o \in R[X]^K$, while for the coinvariants, given an orbit $O$ the corresponding coinvariant is any $o \in O$, standing in for ...

1

Consider the properties of the $dx^i$ compared to the $x^i$. As linear functionals, the basis one-forms $dx^i$ are...well, linear on their arguments. Inner products are bilinear, but we are using two one-forms in each linearly independent term. Notational convention. My opinion? It's very misleading. I would almost always keep the tensor product explicit ...

1

The question was already answered in the comments, but you also asked for a nice proof using universal properties. Assuming rings are defined to be commutative, there is a relatively painless argument using universal properties in the form of natural bijections between Hom-sets. For any $A$-algebra $Y$ we let $Z_f(Y)$ be zero set of $f$ in $Y$. This ...

Only top voted, non community-wiki answers of a minimum length are eligible