3

Note that not all elements of $(U \otimes V) \otimes W$ are simple tensors. This is the problem you are encountering, as your map isn't defined for elements like $(u_1 \otimes v_1) \otimes w_1 + (u_2 \otimes v_2) \otimes w_2$. You can fix this by mapping each simple tensor as above and then extending the map linearly on all elements. This trivially gets ...


3

In general the tensor product does not preserve injective maps. For example, consider $A=\mathbb Z/2\mathbb Z$, $C=\mathbb Z$ and $B=2\mathbb Z$. Then the inclusion induces the zero map $$\mathbb Z/2\mathbb Z\otimes 2\mathbb Z\to\mathbb Z/2\mathbb Z\otimes \mathbb Z$$ since $$\overline a\otimes 2b\mapsto\overline a\otimes 2b=\overline{2a}\otimes b=0\otimes b=...


2

For finite dimensional algebras, flat modules are projective, so the projective dimension of $M$ is the same as its weak dimension, which is $$\sup\{d\mid \text{Tor}^A_d(M,-)\neq0\}.$$ Since the class of left modules $X$ such that $\text{Tor}^A_d(M,X)=0$ is closed under coproducts and extensions, and every module is an iterated extension of coproducts of ...


2

For a finite-dimensional vector space $V$, $V$ is isomorphic to its double dual $V^{**},$ through the natural isomorphism $\varphi:V\rightarrow V^{**}$ given by $\varphi(x)(v)=v(x),$ where $x\in V$ and $v\in V^*.$ Hence, an element of $V$ can be seen as a linear functional that acts on covectors.


2

That is correct (for $\alpha$ an element of the base ring). However, using that as a definition hides the most important property of the tensor product, namely the universal property that it satisfies.


2

$u$ and $v$ are members of some vector space $V$ over a field $F$, and $\phi$ and $\theta$ are members of the corresponding dual vector space $V^*$. So $T$ is a linear map $T:V \times V \times V^* \times V^* \rightarrow F$ such that $T(u,v,\phi,\theta) = \phi(u)\theta(v)$ Since $T$ is a $(2,2)$ tensor, we can write it in component form as $T^{ab}_{cd}$. ...


2

If you like a more algebraic approach, consider the map $$ f : \mathbb Z \to \mathbb Z / m \mathbb Z \otimes_{\mathbb Z} \mathbb Z / n \mathbb Z,\; a \mapsto a \otimes 1. $$ This is clearly surjective, so we must have $$ \mathbb Z / m \mathbb Z \otimes_{\mathbb Z} \mathbb Z / n \mathbb Z \simeq \mathbb Z / \ker f. $$ Now all that is left is to show that $\...


1

Yes, it's the same, in a sense. The property you seek is the existence of a canonical isomorphism between spaces $L(X_1,X_2)\otimes L(Y_1,Y_2)$ and $L(X_1\otimes Y_1,X_2\otimes Y_2)$, for $X_1,X_2,Y_1,Y_2$ being arbitrary vector spaces. This isomorphism $$ \iota : L(X_1,X_2)\otimes L(Y_1,Y_2) \rightarrow L(X_1\otimes Y_1,X_2\otimes Y_2)$$ can be defined by ...


1

As usual, we represent ket $\newcommand{\ket}[1]{\lvert{#1}\rangle}\newcommand{\bra}[1]{\langle{#1}\rvert} \ket{a_1}$ by a column vector $\mathbb{C}^N=\mathbb{C}^{N\times 1}$ and bra $\bra{a_2}$ by a row vector $\mathbb{C}^N=\mathbb{C}^{1\times N}$. The outer product of two kets $\ket{a_1}$ and $\ket{a_2}$ is the ket-bra matrix product $\ket{a_1}\bra{a_2}\...


1

If with respect to a coordinate system $(x^j)$ on $M$ you write $J(\partial_j) = \sum J^i_{~j}\partial_i$, then $J$ can be regarded as $$J = \sum J^i_{~j}\,{\rm d}x^j \otimes \partial_i,$$acting on vector fields by $$X \mapsto J(X) = \sum J^i_{~j}{\rm d}x^j(X)\partial_i.$$There is nothing particular about manifolds or almost-complex structures here. In ...


1

It means that any polynomial $z(s,t)$ can be represented as a sum of polynomials, each of the form $x(s)y(t)$.


1

The indices in the coefficients come from the objects where you evaluate the tensor on. Since $g$ is covariant, it eats vectors. If $(e_1,e_2)$ are dual to $(\epsilon^1, \epsilon^2)$, you can evaluate $g_{ij} = g(e_i,e_j)$, but not $g^{ij} = g(\epsilon^1,\epsilon^2)$, as the latter a priori does not make sense. Then $$\begin{align} g_{11} &= g(e_1,e_1) = ...


1

It probably cant get better than the author of the article answering himself but here a small addition that also answers the question and gives a direct interpretation of this Tor, namely it counts something. Since we do homological algebra, we can assume that the algebra is basic. As in Jeremy Rickards answer one has $Tor_d^A(M,A/radA)=DExt_A^d(M,D(A/radA))$...


1

One short way of getting the desired isomorphism is \begin{align*} (K \otimes A) \otimes_K (K \otimes B) &\cong (A \otimes K) \otimes_K (K \otimes B) \\ &\cong A \otimes (K \otimes_K (K \otimes B)) \tag{$\ast$} \\ &\cong A \otimes K \otimes B \\ &\cong K \otimes A \otimes B \\ &\cong K \otimes (A \otimes B) ...


1

The correct symbol $\otimes$ is produced with \otimes. A strategy is to first find the matrix of $T$ taken with respect to the basis $\mathcal{B}$, and then convert it to the standard basis. With respect to $\mathcal{B}$, it is clear that $$[T]_{\mathcal{B}} = \begin{pmatrix} 0 & 1 \\-1 & 2 \end{pmatrix}.$$Now, we use the tensor transformation law (...


1

You probably meant $\otimes$ instead of $\oplus$. Try using the fact that the matrix elements of a tensor $M=a\otimes b$ in the canonical basis are $M_{ij}=v_i\cdot Mv_j = (a\cdot v_i)(b\cdot v_j)$ where $\cdot$ is the standard dot product and $v_k$ are the basis vectors.


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