Tensor Algebras

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, and to this end we first define tensor algebras, symmetric algebras, and exterior algebras. Throughout, we always regard \(A\) as a commutative ring; in particular, \(A\) then has the IBN property. (§Basis, ⁋Proposition 6)

Definition of Tensor Algebras

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

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

(§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 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 view \(\T(M)\) as an \(\mathbb{N}\)-graded associative unital algebra, we simply replace the left-hand side with the appropriate category. Unpacking this adjunction into a universal property gives 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 \(f=g \circ\iota\).

Moreover, if \(E\) is an \(\mathbb{N}\)-graded \(A\)-algebra and \(u(M)\subseteq E_1\), 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 know that \(\T(u): \T(M) \rightarrow \T(N)\) is surjective.

Properties of Tensor Algebras

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, but to avoid notational complexity we write the target category as \(\Alg{A}\).

First we 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 left adjoint to \(\Hom\) together with a short 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 looks 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:

\(T\left(\bigoplus_{i\in I} M_i\right)\cong \coprod_{i\in I} \T(M_i)\)¹

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

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

and applying the above description yields the following proposition.

Proposition 3 In the above situation, \(\T(M)\) 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 that using the structure constants from §Basis, ⁋Definition 9, we can describe the multiplication in \(\T(M)\); according to the above description, 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 do this, 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 \(\id_{\phi_!M}\). (§Change of Base Ring, ⁋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 following 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 \(\phi_!\T_A(M) \rightarrow \T_B(\phi_!M)\), we can verify that it is the inverse of the above \(B\)-linear map.

Mixed tensor

Now fixing a free \(A\)-module \(M\) and considering the tensor algebra \(\T(M\oplus M^\ast)\) of \(M\oplus M^\ast\), from the result of Proposition 3, \(\T^n(M\oplus M^\ast)\) is

Ah this is ambiguous… it would be neat to define tensor fields using this, but then different orders would be treated as different things… anyway if we resolve this, we just need to explain killing it by applying contraction

Definition of Symmetric Algebras

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 denoted by \(\S(M)\).

From the definition, \(\mathfrak{I}\) is a homogeneous ideal, so it is obvious that \(\T(M)/\mathfrak{I}\) becomes a \(\mathbb{Z}_{\geq 0}\)-graded algebra. Also, 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 \(\S^1(M)\cong M\).

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

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

The product of two elements of \(\S(M)\) is conventionally written as \(xy\), etc.

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

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 \(f=g \circ\iota\).

More generally, fix \(A\)-modules \(M,N\) and a natural number \(n\geq 1\). 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 \(M,N\) and 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 recovers \(\S(u)\).

Properties of Symmetric Algebras

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 \(U:\cAlg{A}\rightarrow \rMod{M}\), so it preserves colimits, and colimits in \(\cAlg{A}\) are given by \(\otimes_A\) (just as coproducts in \(\cRing\) are tensor products). 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.

Let

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

Then 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 \(\S(M)\) is exactly 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 The map \(\S_{B}(B\otimes_AM)\rightarrow B\otimes_A\S_A(M)\) is an isomorphism.

Definition of Exterior Algebras

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 denoted by \(\bigwedge(M)\).

The product of elements in \(\bigwedge(M)\) is conventionally written using \(\wedge\). On the other hand, as in the discussion following 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 \(f=g \circ\iota\).

A property analogous to Proposition 7 holds for exterior algebras as well. For any \(A\)-modules \(M,N\) and 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 \(M,N\) and 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 Exterior Algebras

Similarly, when \(M=\bigoplus_{i\in I} M_i\) is the direct sum of \(A\)-modules \(M_i\), from the fact that \(\bigwedge\) is a left adjoint we know that \(\bigwedge(M)\) must be the colimit of the \(\bigwedge(M_i)\). To make this rigorous, we would need to define colimits in the category of alternating algebras. This comes out as a tensor product, similar to \(\cAlg{A}\), but with the Koszul sign convention attached. This is nothing serious: by definition, when multiplying elements of degrees \(m,n\) in the exterior algebra a sign \((-1)^{mn}\) appears, 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 \(J\subseteq I\), let

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

Then 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 sign simply by swapping the positions of the last two elements, so by giving \(I\) an arbitrary order and arranging accordingly as above, we can avoid meaningless duplication. The following proposition needs even less explanation.

Proposition 14 The map \(\bigwedge_{B}(B\otimes_AM)\rightarrow B\otimes_A\bigwedge_A(M)\) 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 3) On the other hand, in the same post we also verified that the coproduct in the category \(\cRing\) is given by the tensor product \(\otimes\). ↩