\[ \DeclareMathOperator{\End}{End} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\span}{span} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\ad}{ad} \newcommand\d{\mathrm{d}} \newcommand\pref[1]{(\ref{#1})} \]

Section 2.11. Tensor products

Exercise 2.11.2 This problem was skipped because it's too tedious.

Problem 2.11.3

1. A linear map \(f : V \otimes W \to U\) can be associated to the bilinear map \(g(v, w) = f(v \otimes w)\). In the other direction, a bilinear map \(g : V \times W \to U\) can be associated to the linear map defined so that \(f(v \otimes w) = g(v, w)\), extended by linearity to all of \(V \otimes W\). The defining properties of the tensor product space mimic the conditions in the definition of a bilinear map, guaranteeing that this bijection is well-defined.
2. By the definition of the tensor product, the \(\{v_i \otimes w_j\}\) clearly span \(V \otimes W\). To see that they are linearly independent, we use the result from part (a). If \(\sum_{i, j} c_{ij} v_i \otimes w_j = 0\), then for each pair \((i, j)\), let \(g_{ij}\) be the bilinear map such that \(g(v_k, w_l) = 1\) if \(k = i\) and \(l = j\), and 0 otherwise. Let \(f_{ij} : V \otimes W \to k\) be the corresponding linear map from part (a). Then \(f_{ij}(0) = c_{ij}\), so \(c_{ij} = 0\). This is true for all \((i, j)\), so the \(\{v_i \otimes w_j\}\) indeed form a basis.
3. For pure tensors \(t = \varphi \otimes w\), the corresponding homomorphism is defined by \(f(v) = \varphi(v) w\). This is bilinear in \(\varphi\) and \(w\), so it can be uniquely extended to all \(t \in V^* \otimes W\) by linearity. Injectivity follows after choosing a basis for \(V\). If \(t = \sum_i \varphi_i \otimes w_i\) where \(\varphi_i\) is dual to the basis vector \(v_i\), then \(f(v) = \sum_i \varphi_i(v) w_i\), and if the latter vanishes identically, then \(0 = f(v_i) = w_i\) for all \(i\), hence \(t = 0\). Surjectivity also requires us to use the assumption that \(V\) is finite-dimensional; it implies that the \(\varphi_i\) span \(V^*\), so that every \(f \in \Hom(V, W)\) can be written as \(f = \sum_i \varphi_i f(v_i)\), and hence corresponds to the tensor \(t = \sum_i \varphi_i \otimes f(v_i)\).

4. A basis \(B\) for \(S^n V\) is given by all tensors of the form \(v_{i_1} \otimes v_{i_2} \otimes \ldots \otimes v_{i_n}\) where \(i_1 \leq i_2 \leq \ldots \leq i_n\). These span \(S^n V\) because any pure tensor in \(V^{\otimes n}\) differs from one of these basis tensors by a sequence of transpositions. To prove linear independence, we make an observation about the structure of the subspace \(S\) spanned by the \(T - s(T)\). Namely, if \(T = v \otimes w \otimes \ldots\) and the transposition swaps \(v\) and \(w\), then by expanding \(v\), \(w\), and the remaining factors in the basis \(\{v_i\}\), we can write \(T - s(T) = \sum_k t_k - s(t_k)\), where each \(t_k\) is a basis tensor (the product of basis vectors). The general element in \(S\) can then be written in the form \(\sum_m t_m - s_m(t_m)\), where each \(t_m\) is a basis tensor and \(s_m\) is some transposition, since the above argument applies for any transposition. Such a linear combination has the property that for each set of indices \(\{i_1, \ldots, i_n\}\), we collect together terms of the form \(v_{i_{\sigma(1)}} \otimes \ldots \otimes v_{i_{\sigma(n)}}\), where \(\sigma\) is any permutation, the sum of the coefficients of all such terms is zero. A linear combination of the elements in \(B\) cannot have this property since each possible set of indices appears only once, unless all coefficients are zero; hence \(B\) is indeed a basis. The dimension of \(S^n V\) is the size of \(B\), that is, \(\binom{m+n-1}{n}\) (by a stars and bars argument or similar).

   For the \(\wedge^n V\) case, observe that the two tensors \(t_1 = v \otimes w \otimes \ldots\) and \(t_2 = w \otimes v \otimes \ldots\) are additive inverses in \(\wedge^n V\) since their sum equals its own transposition. It follows that a tensor of the form \(t = v_{i_1} \otimes v_{i_2} \otimes \ldots \otimes v_{i_n}\) vanishes in \(\wedge^n V\) if any of the \(i\)'s are equal, and otherwise it can be rearranged so that the \(i\)'s are strictly increasing, at the cost of a possible minus sign. Therefore the set of tensors \(v_{i_1} \otimes \ldots \otimes v_{i_n}\) with \(i_1 < \ldots < i_n\) spans \(\wedge^n V\). Linear independence is proven by a similar argument to that used for \(S^n V\): if \(S\) is the subspace spanned by all pure tensors \(T\) with \(T = s(T)\) for some transposition \(s\), then by writing out the factors of \(T\) in terms of the basis \(\{v_i\}\), we can write the general term of \(S\) in the form \(\sum_m t_m + s_m(t_m)\) where \(t_m\) is a basis tensor and \(s_m\) is a transposition. This then has the property that when we collect terms with some subset of distinct indices \(i_1, \ldots, i_n\), the alternating sum of their coefficients vanishes (where a coefficient is subtracted rather than added if the basis tensor has an odd permutation of the basis vector indices). For a linear combination of the tensors in the set we claim as basis, this cannot be unless all coefficients are zero, since each possible subset of indices appears in only one basis tensor. So our set is indeed a basis. Its size is \(\binom{m}{n}\), the number of subsets of size \(n\) that can be drawn from a set of size \(m\).

5. For this problem, we assume that transposition is defined for all tensors, not just pure tensors, by writing the tensor as a sum of pure tensors and taking the transposition of each term.

   For each pure tensor \(t = w_1 \otimes w_2 \otimes \ldots w_n\) (where the \(w_i\) are not necessarily basis vectors), we can identify its projection down to \(S^n V\) with its symmetrization, \begin{equation*} \mathrm{Sym}(t) = \frac{1}{n!} \sum_\sigma w_{\sigma(1)} \otimes w_{\sigma(2)} \otimes \ldots \otimes w_{\sigma(n)} \end{equation*} where \(\sigma\) ranges over all permutations of \(\{1, \ldots, n\}\). This satisfies \(\mathrm{Sym}(t) = s(\mathrm{Sym}(t))\) for all \(s\), and extending by linearity to all of \(V^{\otimes n}\). This map is well-defined on \(S^n V\) since two tensors differing only by transpositions are mapped to the same symmetrized tensor. To go the other direction, a given \(t \in T\) can simply be identified with its projection down to \(S^n V\); it is easy to see that this is the inverse map.

   For the exterior power, for each \(t = \otimes_i w_i\), we identify its projection down to \(\wedge^n V\) with its antisymmetrization, \begin{equation*} \mathrm{Anti}(t) = \frac{1}{n!} \sum_\sigma \sgn(\sigma) w_{\sigma(1)} \otimes w_{\sigma(2)} \otimes \ldots \otimes w_{\sigma(n)} \end{equation*} This has the desired property that \(\mathrm{Anti}(t) = -s(\mathrm{Anti}(t))\) for all \(s\), and again we can extend it by linearity to all of \(V^{\otimes n}\), and it is well-defined on \(\wedge^n V\) since it annihilates any \(t\) such that \(t = s(t)\) for some transposition \(s\). Again, in the other direction, we simply identify \(t \in T\) with its projection down to \(\wedge^n V\).

   In both cases, the fact that \(k\) has characteristic zero is needed so that division by \(n!\) is always well-defined. We might choose to define Sym and Anti without the factor of \(1/n!\), but this does not solve the problem since if the characteristic divides \(n!\), some nonzero elements of the {symmetric, exterior} power may be mapped to zero.

6. Using the eigenbasis \(\{v_i\}\) (with \(A v_i = \lambda_i v_i\)), each basis tensor \(v_{i_1} \otimes \ldots \otimes v_{i_n}\) is mapped to \(\lambda_{i_1} \cdot \otimes \cdot \lambda_{i_n}\) times itself, so \(\tr(S^n A)\) is the sum of all \(\lambda_{i_1} \cdot \ldots \cdot \lambda_{i_n}\) where \(1 \le i_1 \le \ldots \le i_n \le N\), and likewise \(\tr(\wedge^n A)\) is the sum of all \(\lambda_{i_1} \cdot \ldots \cdot \lambda_{i_n}\) where \(1 \le i_1 < \ldots < i_n \le N\), that is, the \(i_1, \ldots, i_n\) range over all subsets of \(\{1, \ldots, N\}\).

7. Where \(n = N\), \(\tr(\Lambda^N A)\) is therefore the single term \(\lambda_1 \cdot \ldots \cdot \lambda_N\), which is \(\det A\). Also \(\Lambda^N V\) is one-dimensional, so \(\Lambda^N A = (\det A) I\).

   It then follows that \begin{equation*} \det(AB)I = \Lambda^N(AB) = \Lambda^N A \circ \Lambda^N B = (\det A)I \circ (\det B)I = (\det A \det B)I \end{equation*} so \(\det(AB) = \det A \det B\).

Exercise 2.11.5 \(A \otimes_K L\) can be given the structure of an algebra over \(L\) by defining:

1. \(l'(a \otimes_K l) = a \otimes_K (l' l)\)
2. \((a_1 \otimes_K l_1)(a_2 \otimes_K l_2) = (a_1 a_2) \otimes_K (l_1 l_2)\)

and extending by linearity to all of \(A \otimes_K L\). For (1) this is well-defined since the defining relations of the tensor product are annihilated, for example, for all \(a \in A, k \in K, l \in L\): \[ (ka) \otimes_K l - k(a \otimes_K l) \mapsto (ka) \otimes_K (l'l) - k(a \otimes_K (l'l)) \] where the RHS also vanishes by the defining relations of \(\otimes_K\). A similar result holds for the other defining relations. For (2) well-definedness follows by a similar argument. If we take one of the defining relations for, say, the left operand, and use a pure tensor \(a \otimes_K l_2\) as the right operand: \[ (a_1 + a_2) \otimes_K l_1 - a_1 \otimes_K l_1 - a_2 \otimes_K l_2 \mapsto ((a_1 + a_2)a) \otimes_K (l_1 l_2) - (a_1a) \otimes_K (l_1 l_2) - (a_2a) \otimes_K (l_1 l_2) \] which is again a defining relation, and vanishes.

We can easily verify that:

• \(l_1 (l_2 t) = (l_1 l_2) t\)
• \(l (t_1 t_2) = (l t_1) t_2\)
• \((l_1 + l_2) t = l_1 t + l_2 t\)
• \(l(t_1 + t_2) = lt_1 + lt_2\)

where \(l, l_1, l_2 \in L\), \(t, t_1, t_2 \in A \otimes_K L\).

For the second part of the problem, \(V \otimes_K L\) can be given the structure of a module over \(A \otimes_K L\) by defining \begin{equation*} (a \otimes_K l_1)(v \otimes_K l_2) = (av) \otimes_K (l_1 l_2) \end{equation*} and extending by linearity. Proof of well-definedness and the module axioms is very similar so we omit it.

Problem 2.11.6

1. As the problem says, the isomorphism from left to right is given by \((v \otimes_B w) \otimes_C x \mapsto v \otimes_B (w \otimes_C x)\). We can extend this by linearity to the entirety of \((V \otimes_B W) \otimes X\). If we simply choose bases for \(V\), \(W\), and \(X\), it is easy to see that this is well-defined: once we fix \((v_i \otimes_B w_j) \otimes_C x_k \mapsto v_i \otimes_B (w_j \otimes_C x_k)\) for all basis vectors \(v_i \in V, w_j \in W, x_k \in X\), then \((v \otimes_B w) \otimes_C x\) will indeed be mapped to \(v \otimes_B (w \otimes_C x)\) for any \(v \in V, w \in W, x \in X\).

   We need to show that the isomorphism preserves the \((A, D)\)-bimodule structure. Note that \begin{align*} a((v \otimes_B w) \otimes_C x) &= (a(v \otimes_B w)) \otimes_C x \\ &= ((av) \otimes_B w) \otimes_C x \\ &\mapsto (av) \otimes_B (w \otimes_C x) \\ &= a(v \otimes_B (w \otimes_C x)) \end{align*} and a similar fact holds for right-multiplication by \(d \in D\). We can then conclude from linearity that the desired result holds in general.

2. If \(f \mapsto 0\), then \(w \otimes_B v \mapsto f(v)w\) must vanish for all \(w \in W, v \in V\), which is only possible if \(f(v) = 0 \, \forall v \in V\), that is, \(f = 0\). Therefore the homomorphism given in the text is one-to-one.

   We also need it to be onto. Let \(g \in \Hom_C(W \otimes_B V, X)\) be given. We claim that there exists \(f \in \Hom_B(V, \Hom_C(W, X))\) such that \(g(w \otimes_B v) = f(v)w\) for all \(w \in W, v \in V\). To construct such \(f\), notice that if \(v\) is fixed, then \(g(w \otimes_B v)\) is a linear and \(C\)-linear function of \(w\) taking \(W\) into \(X\); let that map be \(f(v)\). If \(v_1, v_2 \in V\), then \(g(w \otimes_B (v_1 + v_2)) = g(w \otimes_B v_1 + w \otimes_B v_2) = g(w \otimes_B v_1) + g(w \otimes_B v_2)\), and similarly \(g(w \otimes_B \alpha v) = \alpha g(w \otimes_B v)\), so \(f\) is linear. Finally, if \(b \in B\), then \(f(bv) = (w \mapsto g(w \otimes_B (bv))) = (w \mapsto g((wb) \otimes_B v)) = b(w \mapsto g(w \otimes_B v))\) according to the left \(B\)-module structure of \(\Hom_C(W, X)\). So \(f\) is \(B\)-linear.

   If \(a \in A\), then \(af \mapsto (w \otimes_B v \mapsto (af)(v)w) = (w \otimes_B v \mapsto f(va)w) = a(w \otimes_B v \mapsto f(v)w)\) since \((w \otimes_B v)a = w \otimes_B (va)\), and hence \(w \otimes_B v\) would be mapped to \(f(va)w\) under \(a(w \otimes_B v \mapsto f(v)w)\). And if \(d \in D\), then \(fd \mapsto (w \otimes_B \mapsto (fd)(v)w) = (w \otimes_B \mapsto (f(v)d)w) = (w \otimes_B \mapsto (f(v)w)d) = (w \otimes_B \mapsto f(v)w)d\). So \(f \mapsto (w \otimes_B v \mapsto f(v)w)\) preserves the \((A, D)\)-bimodule structure.

Problem 2.11.7 For \(a \in A\), define \(a(m \otimes_A n) \equiv (am) \otimes_A n = m \otimes_A (an)\) for the pure tensors, and otherwise, if \(t \in M \otimes_A N\) is expressed as \(t = \sum_i m_i \otimes_A n_i\), then \(at = \sum_i (am_i) \otimes_A n_i\). To get well-definedness, note that defining relations are mapped to zero, for example, \(a((m_1 \otimes_A n) + (m_2 \otimes_A n) - ((m_1 + m_2) \otimes_A n) = a(m_1 \otimes_A n) + a(m_2 \otimes_A n) - a((m_1 + m_2) \otimes_A n) = (am_1 \otimes_A n) + (am_2 \otimes_A n) - (a(m_1 + m_2) \otimes_A n) = 0\) since the last expression is in the form of a defining relation. It is easy to verify that the module axioms are satisfied.