Dual Spaces

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Module \(\Hom_\lMod{A}(M,N)\)

Let \(M\) and \(N\) be arbitrary left \(A\)-modules. Then \(\Hom_\lMod{A}(M,N)\) is an abelian group, but it does not generally carry the structure of a left \(A\)-module. That is, in general, for arbitrary \(\alpha\in A\) and \(u\in\Hom_\lMod{A}(M,N)\), the map

\[x\mapsto \alpha u(x)\]

\(\alpha u: M \rightarrow N\) defined by the above formula is not an \(A\)-linear map. This can be seen from the formula

\[(\alpha u)(\beta x)=\alpha u(\beta x)=\alpha \beta u(x)\neq \beta\alpha u(x)=\beta\cdot ((\alpha u)(x))\]

for arbitrary \(\beta\in A\) and \(x\in M\). However, from this formula we also see that if \(\alpha\) lies in the center of \(A\), then \(\alpha u\) becomes an \(A\)-linear map. That is, with respect to the center \(Z(A)\) of \(A\), the set \(\Hom_\lMod{A}(M,N)\) is a left \(Z(A)\)-module. By similar reasoning, for arbitrary right \(A\)-modules \(M,N\), we see that \(\Hom_\rMod{A}(M,N)\) is a right \(Z(A)\)-module.

Now suppose \(M\) and \(N\) are left \(A\)-modules, and in particular that \(N\) carries a right \(B\)-module structure compatible with this, i.e., \(N\) is an \((A,B)\)-bimodule. Then for any \(\beta\in B\) and \(u\in\Hom_\lMod{A}(M,N)\), the map defined by

\[x\mapsto u(x)\beta\]

\(u\beta: M \rightarrow N\) is an \(A\)-linear map, as can be seen from the formula

\[(u\beta)(\alpha x)=u(\alpha x)\beta=\alpha u(x)\beta=\alpha((u\beta)(x))\]

The same reasoning applies to a right \(A\)-module \(M\) and a \((B,A)\)-bimodule \(N\).

Definition of Dual Spaces

Any ring \(A\) has a natural \((A,A)\)-bimodule structure given by its multiplication. Therefore, by the preceding argument, we may regard \(\Hom_{\lMod{A}}(M, A)\) as a right \(A\)-module.

Definition 1 The right \(A\)-module \(\Hom_{\lMod{A}}(M, A)\) defined above is called the dual module of \(M\), and is denoted by \(M^\ast\).

Similarly, given a right \(A\)-module \(M\), we may view \(\Hom_{\rMod{A}}(M,A)\) as a left \(A\)-module, and we call it the dual module of \(M\). In the special case \(M=A\), to avoid confusion we write \(A_l\) for \(A\) regarded as a left \(A\)-module and \(A_r\) for \(A\) regarded as a right \(A\)-module; then one can verify the two equalities \(A_l^\ast=A_r\) and \(A_r^\ast=A_l\).

By definition, for any \(x\in M\) and \(\xi\in M^\ast\), the pair assigns an element \(\xi(x)\in A\). We write this as \(\langle x, \xi\rangle\), and call this notation the Kronecker pairing.

Definition 2 For any \(A\)-module \(M\) and its dual \(M^\ast\), elements \(x\in M\) and \(\xi\in M^\ast\) are said to be orthogonal if \(\langle x,\xi\rangle=0\).

If every pair of elements from a subset of \(M\) and a subset of \(M^\ast\) is orthogonal, we say that the two subsets are orthogonal. Now fix an arbitrary \(x\in M\), and let \(\xi,\xi_1,\xi_2\in M^\ast\) and \(\alpha\in A\). Then

\[\langle x, \xi_1+\xi_2\rangle=\langle x, \xi_1\rangle,+\langle x,\xi_2\rangle=0,\qquad \langle x,\xi\cdot\alpha\rangle=\langle x,\xi\rangle\alpha=0\]

hence, for a fixed subset \(S\) of \(M\), the collection of elements of \(M^\ast\) orthogonal to the elements of \(S\) forms a submodule of \(M^\ast\).

Definition 3 The submodule of \(M^\ast\) defined as above is called the submodule orthogonal to \(S\), and is denoted by \(S^\perp\).

For an arbitrary subset \(T\subseteq M^\ast\), we can similarly define \(T^\perp\) by the formula

\[T^\perp=\{x\in M\mid \langle x, \xi\rangle=0\text{ for all $\xi\in T$}\}\]

but note that here \(T^\perp\) is defined as a submodule of \(M\), not of \(M^{\ast\ast}\).

Transpose of Linear Maps

Let \(u:M \rightarrow N\) be an arbitrary \(A\)-linear map. Then the abelian group homomorphism given in [Algebraic Structures] §Modules, ⁋Proposition 8

\[\Hom(u,A):\Hom_{\lMod{A}}(N,A)\rightarrow\Hom_{\lMod{A}}(M,A)\]

is compatible with the right action of \(A\). That is, \(\Hom(u,A)\) is a right \(A\)-module homomorphism.

Definition 4 For an \(A\)-linear map \(u:M \rightarrow N\) between left \(A\)-modules, the right \(A\)-module homomorphism defined above is called the transpose of \(u\), and is denoted by \(u^\ast\).

\(u^\ast\) is determined by its values \(u^\ast(\xi)\in M^\ast\) at any \(\xi\in N^\ast\), and in turn each \(u^\ast(\xi)\in M^\ast\) is determined by its values at any \(x\in M\)

\[u^\ast(\xi)(x)=\langle x, u^\ast(\xi)\rangle\]

By definition of \(u^\ast=\Hom(u,A)\), we have \(u^\ast(\xi)=\xi\circ u\). Thus the above formula can be written as

\[\langle u(x),\xi\rangle=\langle x, u^\ast\xi\rangle\]

and conversely, if this formula holds for all \(x\in M\) and all \(\xi\in N^\ast\), then \(u^\ast\) is uniquely determined.

Moreover, by the functoriality of \(\Hom(-,A)\) and [Algebraic Structures] §Modules, ⁋Proposition 8, we obtain the following proposition.

Proposition 5 The following hold.

1. For any two \(A\)-linear maps \(u,v:M \rightarrow N\), we have \((u+v)^\ast=u^\ast+v^\ast\).
2. For any two \(A\)-linear maps \(u:M \rightarrow N\) and \(v:N \rightarrow L\), we have \((v\circ u)^\ast=u^\ast\circ v^\ast\).
3. For any \(M\), we have \((\id_M)^\ast=\id_{M^\ast}\).
4. For any \(A\)-linear isomorphism \(u:M \rightarrow N\), we have \((u^{-1})^\ast=(u^\ast)^{-1}\).

Dual Basis

Suppose the \(A\)-module \(M\) has a basis \((x_i)_{i\in I}\). (§Basis, ⁋Definition 1) That is, there exists an isomorphism

\[\varepsilon: A^{\oplus I} \rightarrow M\]

Taking the dual of this isomorphism yields an isomorphism of right \(A\)-modules

\[\varepsilon^\ast: M^\ast \rightarrow (A_l^{\oplus I})^\ast=\Hom_{\lMod{A}}(A_l^{\oplus I}, A_l)\cong \prod_{i\in I}\left(\Hom_\lMod{A}(A_l,A_l)\right)\cong \prod_{i\in I} A_r\]

Now consider the elements on the right-hand side whose \(i\)th component is \(1\) and whose remaining components are \(0\), and denote the preimage of each such element under \(\varepsilon^\ast\) by \(e_i^\ast\). Then we see that the formula

\[\langle e_i, e_j^\ast\rangle=\delta_{ij}\]

holds. The collection of these elements is linearly independent, but if \(I\) is infinite it does not form a basis of \(M^\ast\). However, if \(I\) is finite then \(\prod_{i\in I} A\cong \bigoplus_{i\in I}A\), so they constitute a basis.

Definition 6 Fix an arbitrary free module \(M\) and a basis \((e_i)_{i\in I}\). Then the family \((e_i^\ast)_{i\in I}\) of elements of \(M^\ast\) defined above is called the coordinate form corresponding to \((e_i)_{i\in I}\).
If \(M\) is a finitely generated free module, then this family \((e_i^\ast)_{i\in I}\) is a basis of \(M^\ast\), and is called the dual basis of \((e_i)\).

Bidual Space

For any left \(A\)-module \(M\), the dual \(M^\ast\) of \(M\) is a right \(A\)-module, and the dual \(M^{\ast\ast}\) of \(M^\ast\) is again a left \(A\)-module. Now, for any \(x\in M\), one can verify that the map defined by the formula

\[\langle x,-\rangle: M^\ast \rightarrow A\]

is a right \(A\)-module homomorphism. Thus the above formula defines a map from \(M\) to \(M^{\ast\ast}\), and one can verify that this map is also linear. In general, this map is neither injective nor surjective.

Definition 7 If the above map \(M \rightarrow M^{\ast\ast}\) is bijective, then \(M\) is called reflexive.

Then the following holds.

Proposition 8 For any free module \(M\), the map \(M \rightarrow M^{\ast\ast}\) defined above is injective. If in addition \(M\) is finitely generated, then this map is bijective.