Tensor Algebra

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We now define the determinant; to this end, we first define the tensor algebra, the symmetric algebra, and the exterior algebra. Throughout, \(A\) will always be taken to be a commutative ring. In particular, \(A\) then has the IBN property. (§Bases, ⁋Proposition 6)

Definition of the Tensor Algebra

For any \(A\)-module \(M\), we previously defined the free algebra \(F(M)\) generated by \(M\) via the formula

\[F(M)=\bigoplus_{n\geq 0} M^{\otimes n}\]

([Algebraic Structures] §Algebras, ⁋Proposition 4). This is not merely an algebra, but naturally carries the structure of an \(\mathbb{N}_{\geq 0}\)-graded associative unital algebra. We name it as follows.

Definition 1 The \(F(M)\) defined above is called the tensor algebra of \(M\), and is denoted by \(\T(M)\).

We shall write each component \(M^{\otimes n}\) as \(\T^n(M)\). Then since \(\T^1(M)=M\), there is a canonical injection \(\iota: M \rightarrow \T(M)\).

Now, considering the adjunction \(T\dashv U\), the map \(\iota\) is the image of \(\id_{\T(M)}\) under the adjunction

\[\Hom_{\Alg{A}}(\T(M), \T(M))\cong \Hom_{\rMod{A}}(M, U\T(M))\]

and if we regard \(\T(M)\) as an \(\mathbb{N}\)-graded associative unital algebra, we simply replace the left-hand side with the appropriate category. Unpacking this adjoint into a universal property yields the following.

Proposition 2 Let an \(A\)-algebra \(E\) and an \(A\)-linear map \(u:M \rightarrow E\) be given. Then there exists a unique \(A\)-algebra homomorphism \(g: \T(M) \rightarrow E\) such that \(u=g \circ\iota\).

Moreover, if \(E\) is an \(\mathbb{N}\)-graded \(A\)-algebra and \(u(M)\subseteq E_1\) holds, then the \(A\)-algebra homomorphism \(g\) obtained above is an \(\mathbb{N}\)-graded \(A\)-algebra homomorphism.

If the linear map \(u\) above is surjective, then since \(\T(N)\) is generated by \(\T^1(N)\), we see that \(\T(u): \T(M) \rightarrow \T(N)\) is surjective.

Properties of the Tensor Algebra

We now examine how operations in \(\rMod{A}\) behave when transported via the functor \(T:\rMod{A} \rightarrow \Alg{A}\). We are particularly interested in direct sums and extension of scalars. The discussion in this section, as in Proposition 2, remains valid if we understand \(T\) as a functor from \(\rMod{A}\) to the category of associative unital \(\mathbb{N}\)-graded \(A\)-algebras; however, to avoid notational complexity, we shall write the target category as \(\Alg{A}\).

First, consider the case of direct sums. Let \(M=\bigoplus_{i\in I} M_i\) be the direct sum of \(A\)-modules \(M_i\). Then, using the fact that \(\otimes\) is a left adjoint of \(\Hom\) and a little induction, we obtain the isomorphism

\[\bigoplus_{(i_1,\ldots, i_n)\in I^n}M_{i_1}\otimes\cdots\otimes M_{i_n}\cong \T^n(M)\]

and thus \(\T(M)\) is given as the direct sum

\[\T(M)\cong\bigoplus_{n\geq 0} \T^n(M)\cong\bigoplus_{n\geq 0}\bigoplus_{(i_1,\ldots, i_n)\in I^n}M_{i_1}\otimes\cdots\otimes M_{i_n}\]

This may look complicated as a formula, but it is essentially nothing more than unpacking the coproduct of graded algebras on the right-hand side, since \(T\) is a left adjoint:¹

\[T\left(\bigoplus_{i\in I} M_i\right)\cong \coprod_{i\in I} \T(M_i)\]

In particular, for any free \(A\)-module \(M\), let \(\mathcal{B}=(e_i)_{i\in I}\) be a basis of \(M\). Then

\[M=\bigoplus_{i\in I} Ae_i\]

and applying the above explanation yields the following proposition.

Proposition 3 In the situation above, \(\T(M)\) has as its basis the elements \(e_s\) of the form

\[e_s=e_{i_1}\otimes\cdots\otimes e_{i_n},\qquad\text{$s$ a finite sequence $(i_1,i_2,\ldots,i_n)$ in $I$}\]

This is because each \(\T^n(M)\) has as its basis the \(e_s\) defined using finite sequences \(s\) of length \(n\), and their direct sum is \(\T(M)\). On the other hand, we know from §Bases, ⁋Definition 9 that the structure constants can be used to describe the multiplication in \(\T(M)\); according to the above explanation, this is nothing other than concatenation of sequences. That is, for two sequences

\[s=(i_1,\ldots, i_m),\qquad t=(j_1,\ldots, j_n)\]

if we define \(st\) as the sequence

\[st=(i_1,\ldots, i_m,j_1,\ldots, j_n)\]

then the equation defining the structure constant becomes

\[e_se_t=e_{st}\]

In the case of extension of scalars, let a ring homomorphism \(\phi: A \rightarrow B\) be given, and let \(M\) be an \(A\)-module. Then there exist the extension of scalars \(\phi_!: \rMod{A} \rightarrow\rMod{B}\) and the two functors \(\T_A: \rMod{A} \rightarrow \Alg{A}\), \(\T_B:\rMod{B} \rightarrow \Alg{B}\), and \(\phi_!:\Alg{A} \rightarrow\Alg{B}\) is also defined in the obvious way. Through this, we obtain the following (graded) \(B\)-linear map:

Proposition 4 The \(B\)-linear map \(\T_{B}(B\otimes_AM)\rightarrow B\otimes_A\T_A(M)\) obtained above is an isomorphism.

Proof

It suffices to construct an inverse. To this end, first from the adjunction

\[\Hom_\rMod{B}(\phi_!M,\phi_!M)\cong\Hom_\rMod{A}(M, \phi^\ast \phi_!M)\]

let us obtain the \(A\)-linear map \(i: M \rightarrow \phi^\ast\phi_!M\) corresponding to \(\id_{\phi_!M}\). ([Algebraic Structures] §Change of Scalars, ⁋Proposition 6) Then, viewing the \(A\)-module \(\phi^\ast\phi_!M\) as the \(B\)-module \(\phi_!M\) and considering

\[\iota_{\phi_!M}: \phi_!M \rightarrow \T_B(\phi_!M)\]

this is an \(A\)-linear map from the \(A\)-module \(M\) to the \(A\)-module \(\phi^\ast \T_B(\phi_!M)\) (more precisely, \(U\phi^\ast \T_B(\phi^\ast\phi_!M)\)). Therefore, by Proposition 2, there exists a unique \(A\)-algebra homomorphism \(T_A(M)\rightarrow \phi^\ast T_{B}(\phi_!M)\) making the following diagram

commute. Now, via the adjunction

\[\Hom_{\Alg{A}}(\T_A(M), \phi^\ast \T_B(\phi_!M))\cong \Hom_\Alg{B}(\phi_! \T_A(M), \T_B(\phi_!M))\]

if we view this as a \(B\)-linear map \(\phi_!\T_A(M) \rightarrow \T_B(\phi_!M)\), one can verify that this becomes the inverse of the \(B\)-linear map above.

Mixed Tensors

Now recall an \(A\)-module \(M\), its dual module \(M^\ast\), and the Kronecker pairing \(\langle x,\xi\rangle\) between them. (§Dual Spaces, ⁋Definition 1) Many objects treated in linear algebra can be found inside tensor products of several copies of \(M\) and \(M^\ast\); for example, if \(M\) is finitely generated projective, then by §Hom and the Tensor Product, ⁋Corollary 4 we have \(M^\ast\otimes_AM\cong \End_\rMod{A}(M)\). To treat such objects all at once, consider the tensor algebra \(\T(M\oplus M^\ast)\); applying the direct sum decomposition examined in the previous section to \(M_1=M\), \(M_2=M^\ast\), we obtain the isomorphism

\[\T^n(M\oplus M^\ast)\cong\bigoplus_{(i_1,\ldots, i_n)\in\{1,2\}^n} M_{i_1}\otimes\cdots\otimes M_{i_n}\]

That is, each summand of \(\T^n(M\oplus M^\ast)\) is a tensor product of length \(n\) made from \(M\) and \(M^\ast\), and even if the numbers of \(M\) and \(M^\ast\) contained are the same, they are treated as different summands if their arrangement order differs. For example, when \(n=2\), \(M\otimes_AM^\ast\) and \(M^\ast\otimes_AM\) are different summands in the above decomposition.

However, this distinction carries no information beyond notation. Suppose \((i_1,\ldots, i_n)\) points to \(M\) at \(p\) positions and to \(M^\ast\) at \(q=n-p\) positions, and for \((z_1,\ldots, z_n)\in M_{i_1}\times\cdots\times M_{i_n}\), write the components belonging to \(M\) in their original order as \(x_1,\ldots, x_p\), and the components belonging to \(M^\ast\) in their original order as \(\xi_1,\ldots, \xi_q\). Then the function defined by the formula

\[(z_1,\ldots, z_n)\mapsto x_1\otimes\cdots\otimes x_p\otimes\xi_1\otimes\cdots\otimes\xi_q\]

is \(A\)-linear in each component, so by the universal property of the tensor product, an \(A\)-linear map \(M_{i_1}\otimes\cdots\otimes M_{i_n}\rightarrow M^{\otimes p}\otimes_A(M^\ast)^{\otimes q}\) is induced. The map in the opposite direction is obtained in the same way, and these two are inverses of each other on decomposable tensors; since decomposable tensors generate the whole space, this map is an isomorphism. Thus each summand is canonically isomorphic to

\[\T^p_q(M)=M^{\otimes p}\otimes_A (M^\ast)^{\otimes q}\]

with the components from \(M\) gathered at the front, and the distinction of arrangement order may be erased without loss of information. We call the elements of \(\T^p_q(M)\) tensors of contravariant order \(p\) and covariant order \(q\), or simply tensors of type \((p,q)\), and in particular when \(p,q\geq 1\) we call them mixed tensors. These names are classical terms originating from the way the coordinates of each component transform under a change of basis. By definition, \(\T^p_0(M)=\T^p(M)\), \(\T^0_q(M)=\T^q(M^\ast)\), and \(\T^0_0(M)=A\).

What distinguishes a mixed tensor from a pure tensor power is that components of \(M\) and \(M^\ast\) coexist within a single tensor, and therefore there is an operation that pairs and cancels them via the Kronecker pairing. Fix \(p,q\geq 1\) and \(1\leq i\leq p\), \(1\leq j\leq q\), and define the function \(M^p\times (M^\ast)^q \rightarrow \T^{p-1}_{q-1}(M)\) by the formula

\[(x_1,\ldots, x_p,\xi_1,\ldots, \xi_q)\mapsto \langle x_i,\xi_j\rangle\cdot x_1\otimes\cdots\otimes x_{i-1}\otimes x_{i+1}\otimes\cdots\otimes x_p\otimes \xi_1\otimes\cdots\otimes \xi_{j-1}\otimes\xi_{j+1}\otimes\cdots\otimes \xi_q\]

Since \(A\) is commutative, the Kronecker pairing is \(A\)-bilinear, and the remaining components are simply placed into the tensor product, so this function is \(A\)-linear in each component. Therefore, as above, a unique \(A\)-linear map

\[c^i_j: \T^p_q(M)\rightarrow \T^{p-1}_{q-1}(M)\]

is induced, and we call this the contraction of the \(i\)-th contravariant component and the \(j\)-th covariant component. As its name suggests, contraction is the operation that reduces the tensor type from \((p,q)\) to \((p-1,q-1)\).

In the simplest case \(p=q=1\), the map \(c^1_1: M\otimes_AM^\ast\rightarrow A\) sends \(x\otimes\xi\) to \(\langle x,\xi\rangle\), and viewing this through the canonical isomorphism \(M\otimes_AM^\ast\cong M^\ast\otimes_AM\) that swaps the two components, it coincides with the \(A\)-linear map \(\tau: M^\ast\otimes_AM \rightarrow A\) defined in §Hom and the Tensor Product, §§Trace. In particular, if \(M\) is finitely generated projective, then under the isomorphism \(M^\ast\otimes_AM\cong\End_\rMod{A}(M)\) mentioned above, \(c^1_1\) becomes exactly the trace map of §Hom and the Tensor Product, ⁋Definition 6. In this sense, the general contraction \(c^i_j\) can be thought of as the operation that takes the trace on the type \((1,1)\) part formed by the \(i\)-th contravariant component and the \(j\)-th covariant component, while leaving the remaining components unchanged.

Let us also examine this in coordinates. Suppose \(M\) is a finitely generated free \(A\)-module, and fix a basis \((e_k)_{1\leq k\leq r}\) of \(M\) and its dual basis \((e_k^\ast)_{1\leq k\leq r}\). (§Dual Spaces, ⁋Definition 6) Then by the same argument as in Proposition 3, \(\T^p_q(M)\) is a free \(A\)-module with basis the elements of the form

\[e_{s_1}\otimes\cdots\otimes e_{s_p}\otimes e_{t_1}^\ast\otimes\cdots\otimes e_{t_q}^\ast\]

For example, expanding a type \((1,1)\) tensor \(z\in M\otimes_AM^\ast\) in this basis and writing

\[z=\sum_{k,l=1}^r a^k_l (e_k\otimes e_l^\ast)\]

since \(\langle e_k, e_l^\ast\rangle=\delta_{kl}\), we obtain the formula

\[c^1_1(z)=\sum_{k,l=1}^r a^k_l\langle e_k, e_l^\ast\rangle=\sum_{k=1}^r a^k_k\]

which computes the trace of the endomorphism corresponding to \(z\) as the sum of the diagonal entries of the coefficient matrix \((a^k_l)\). The general \(c^i_j\) is similarly the operation of setting the \(i\)-th upper index and the \(j\)-th lower index to the same value and summing over that value, which corresponds to the classical tensor notation convention of pairing and canceling indices that appear repeatedly above and below. Finally, applying contraction \(p\) times repeatedly to a mixed tensor with \(p=q\) yields an element of \(\T^0_0(M)=A\), that is, a scalar.

Definition of the Symmetric Algebra

Definition 5 For any \(A\)-module \(M\), consider the two-sided ideal of the tensor algebra \(\T(M)\)

\[\mathfrak{I}=\langle x\otimes y-y\otimes x\mid x,y\in M\rangle\]

Then the quotient algebra \(\T(M)/\mathfrak{I}\) is called the symmetric algebra of \(M\), and is written \(\S(M)\).

By definition, since \(\mathfrak{I}\) is a homogeneous ideal, it is obvious that \(\T(M)/\mathfrak{I}\) becomes a \(\mathbb{Z}_{\geq 0}\)-graded algebra. Moreover, since each generator \(x\otimes y-y\otimes x\) is an element of degree \(2\), taking the quotient by \(\mathfrak{I}\) has no effect on \(\T^0(M)\) and \(\T^1(M)\). That is, \(\S^0(M)\cong A\) and \(\S^1(M)\cong M\).

That \(\S(M)\) is a commutative unital associative algebra is obvious from the definition. This is because \(\S(M)\) is generated by the elements of \(\S^1(M)\), and for any \(x,y\in \S^1(M)\cong M\) we have

\[x\otimes y\equiv y\otimes x\pmod{\mathfrak{I}}\]

It is customary to write the product of two elements of \(\S(M)\) in the multiplicative notation \(xy\), etc.

On the other hand, the following universal property is also obtained immediately from the universal property of the quotient algebra and Proposition 2.

Proposition 6 Let an \(A\)-algebra \(E\) and an \(A\)-linear map \(u:M \rightarrow E\) satisfying the condition

\[u(x)u(y)=u(y)u(x)\qquad\text{for all $x,y\in M$}\]

be given. Then there exists a unique \(A\)-algebra homomorphism \(g: \S(M) \rightarrow E\) such that \(u=g \circ\iota\).

More generally, let arbitrary \(A\)-modules \(M,N\) and a natural number \(n\geq 1\) be fixed. A symmetric linear map from \(M^n\) to \(N\) is one satisfying the condition

\[f(x_{\sigma(1)},x_{\sigma(2)},\ldots, x_{\sigma(n)})=f(x_1,x_2,\ldots, x_n),\qquad \sigma\in S_n\]

for all \((x_i)\in M^n\) and \(\sigma\in S_n\).

Proposition 7 For two \(A\)-modules \(M,N\), for any \(A\)-linear map \(g:\S^n(M) \rightarrow N\), the function defined by the formula

\[(x_1,x_2,\ldots, x_n) \mapsto g(x_1x_2\cdots x_n)\]

is \(n\)-linear, and through this correspondence a bijective \(A\)-module homomorphism is defined from \(\Hom_{\lMod{A}}(\S^n(M), N)\) to the \(A\)-module of symmetric \(n\)-linear maps \(M^n \rightarrow N\).

Here, the \(A\)-module \(\S^n(M)\) is called the \(n\)-th symmetric power of \(M\). Then for any \(A\)-linear map \(u:M \rightarrow N\), \(\S^n(u): \S^n(M) \rightarrow \S^n(N)\) is induced, and taking their direct sum allows us to recover \(\S(u)\).

Properties of the Symmetric Algebra

Earlier we examined how the functor \(T\) and operations in \(\rMod{A}\) behave. We now prove that these results also hold for \(S\).

First, let \(M=\bigoplus_{i\in I} M_i\) be the direct sum of \(A\)-modules \(M_i\). Then we obtain the isomorphism

\[\S(M)\cong \bigotimes_{i\in I} \S(M_i)\]

This is because \(S\) is the left adjoint of the forgetful functor \(U:\cAlg{A}\rightarrow \rMod{A}\), so it preserves colimits, and the colimit in \(\cAlg{A}\) is given by \(\otimes_A\) (just as the coproduct in \(\cRing\) is the tensor product). In particular, as in Proposition 3, once we fix a basis \((e_i)\) of a free \(A\)-module, we obtain the following proposition.

Proposition 8 For a free \(A\)-module \(M\) and its basis \((e_i)_{i\in I}\), let \(\alpha:I \rightarrow \mathbb{N}\) be a finitely supported function.

Writing

\[e^\alpha=\prod_{i\in I} e_i^{\alpha(i)}\]

the collection of all such elements forms a basis of \(\S(M)\).

Their multiplication is given by \(e^\alpha e^\beta=e^{\alpha+\beta}\). That is, in this case \(\S(M)\) becomes precisely the polynomial algebra \(A[\x_i]_{i\in I}\).

The result corresponding to Proposition 4 is the following proposition, and its proof is identical as well.

Proposition 9 \(\S_{B}(B\otimes_AM)\rightarrow B\otimes_A\S_A(M)\) is an isomorphism.

Definition of the Exterior Algebra

Definition 10 For any \(A\)-module \(M\), consider the two-sided ideal of the tensor algebra \(\T(M)\)

\[\mathfrak{J}=\langle x\otimes x\mid x\in M\rangle\]

Then the quotient algebra \(\T(M)/\mathfrak{J}\) is called the exterior algebra of \(M\), and is written \(\bigwedge(M)\).

It is customary to write the product of elements in \(\bigwedge(M)\) as \(\wedge\). On the other hand, by the same discussion as after Definition 5, it is obvious that \(\mathfrak{J}\) is a homogeneous ideal and that the canonical inclusion \(\iota:M \hookrightarrow\bigwedge(M)\) exists. Also, for the same reason as in Proposition 6, the following universal property holds.

Proposition 11 Let an \(A\)-algebra \(E\) and an \(A\)-linear map \(u:M \rightarrow E\) satisfying the condition

\[u(x)^2=0\qquad\text{for all $x\in M$}\]

be given. Then there exists a unique \(A\)-algebra homomorphism \(g: \bigwedge(M) \rightarrow E\) such that \(u=g \circ\iota\).

A property similar to Proposition 7 also holds for the exterior algebra. In what follows we assume that the characteristic of \(A\) is not \(2\). For arbitrary \(A\)-modules \(M,N\) and an integer \(n\geq 1\), an alternating linear map \(f\) from \(M^n\) to \(N\) is one satisfying the condition

\[f(x_{\sigma(1)},x_{\sigma(2)},\ldots, x_{\sigma(n)})=\epsilon(\sigma)f(x_1,x_2,\ldots, x_n),\qquad \sigma\in S_n\]

for all \((x_i)\in M^n\) and \(\sigma\in S_n\). This is equivalent to the condition that for any \(x_1,\ldots, x_{n-1}\) and \(x\),

\[f(x_1,\ldots, x_i, x,x,x_{i+1},\ldots, x_{n-1})=0\]

holds.

Proposition 12 For two \(A\)-modules \(M,N\), for any \(A\)-linear map \(g:\bigwedge^n(M) \rightarrow N\), the function defined by the formula

\[(x_1,x_2,\ldots, x_n) \mapsto g(x_1\wedge x_2\wedge\cdots\wedge x_n)\]

is \(n\)-linear, and through this correspondence a bijective \(A\)-module homomorphism is defined from \(\Hom_{\lMod{A}}(\bigwedge^n(M), N)\) to the \(A\)-module of alternating \(n\)-linear maps \(M^n \rightarrow N\).

Properties of the Exterior Algebra

Likewise, when \(M=\bigoplus_{i\in I} M_i\) is the direct sum of \(A\)-modules \(M_i\), we know from \(\bigwedge\) being a left adjoint that \(\bigwedge(M)\) must be the colimit of the \(\bigwedge(M_i)\). To make this rigorous, one must define the colimit in the category of alternating algebras. This arises similarly to \(\cAlg{A}\) via the tensor product, but with the Koszul sign convention attached. This is nothing serious: by definition, multiplying elements of degrees \(m,n\) in the exterior algebra produces a sign \((-1)^{mn}\), and this simply reflects that. We have no immediate need to write this out rigorously, so we only introduce the following proposition.

Proposition 13 For a free \(A\)-module \(M\) and its basis \((e_i)_{i\in I}\), fix a total ordering on \(I\). For any finite subset \(J\subseteq I\), writing

\[e_J=e_{j_1}\wedge e_{j_2}\wedge\cdots\wedge e_{j_k},\qquad j_1<\cdots < j_k, \quad J=\{j_1,\ldots, j_k\}\]

the collection of all such \(e_J\) forms a basis of \(\bigwedge (M)\).

For example, \(e_1\wedge e_2\wedge e_3\) and \(e_1\wedge e_3\wedge e_2\) become the same element up to a sign difference simply by swapping the last two elements, so as above one can avoid meaningless duplication by giving \(I\) an arbitrary order and arranging accordingly. The following proposition needs even less explanation.

Proposition 14 \(\bigwedge_{B}(B\otimes_AM)\rightarrow B\otimes_A\bigwedge_A(M)\) is an isomorphism.

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.

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1. Recall that the coproduct in the category \(\Ring\) was defined in a manner similar to the free product. (§Products, Coproducts, and Tensor Products of Rings, ⁋Proposition 4) On the other hand, in the same post we also verified that the coproduct in the category \(\cRing\) is given by the tensor product \(\otimes\). ↩