Dual Spaces

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.

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

Fix arbitrary left \(A\)-modules \(M\) and \(N\). Then \(\Hom_\lMod{A}(M,N)\) is an abelian group, but in general it does not carry the structure of a left \(A\)-module. That is, for arbitrary \(\alpha\in A\) and \(u\in\Hom_\lMod{A}(M,N)\), the function \(\alpha u: M \rightarrow N\) defined by

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

is not an \(A\)-linear map. This can be seen from the following computation for arbitrary \(\beta\in A\) and \(x\in M\):

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

However, this very computation also shows that if \(\alpha\) lies in the center of \(A\), then \(\alpha u\) is an \(A\)-linear map. Thus, for the center \(Z(A)\) of \(A\), the abelian group \(\Hom_\lMod{A}(M,N)\) is a left \(Z(A)\)-module. By the same reasoning, for arbitrary right \(A\)-modules \(M\) and \(N\), the abelian group \(\Hom_\rMod{A}(M,N)\) is a right \(Z(A)\)-module.

On the other hand, suppose left \(A\)-modules \(M\) and \(N\) are given, and in particular \(N\) carries a right \(B\)-module structure compatible with the given one, i.e. \(N\) is an \((A,B)\)-bimodule. Then for arbitrary \(\beta\in B\) and \(u\in\Hom_\lMod{A}(M,N)\), the function \(u\beta: M \rightarrow N\) defined by

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

is an \(A\)-linear map, as can be seen from the following computation:

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

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

Definition of Dual Spaces

Any ring \(A\) carries a natural \((A,A)\)-bimodule structure via the multiplication defined on it. 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 \(M^\ast\).

Similarly, if a right \(A\)-module \(M\) is given, then \(\Hom_{\rMod{A}}(M,A)\) can be viewed as a left \(A\)-module, and we also call this 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 identities \(A_l^\ast=A_r\) and \(A_r^\ast=A_l\).

By definition, for arbitrary \(x\in M\) and \(\xi\in M^\ast\), the pair \((x,\xi)\) specifies 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 \(M^\ast\), we say that \(x\in M\) and \(\xi\in M^\ast\) are orthogonal if \(\langle x,\xi\rangle=0\).

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

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

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

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

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

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

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

Transpose of a Linear Map

Let an arbitrary \(A\)-linear map \(u:M \rightarrow N\) be given. Then the abelian group homomorphism

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

from [Algebraic Structures] §Modules, ⁋Proposition 8 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 \(u^t\).

The map \(u^t\) is determined by its value \(u^t(\xi)\in M^\ast\) at arbitrary \(\xi\in N^\ast\), and in turn \(u^t(\xi)\in M^\ast\) is determined by its value at arbitrary \(x\in M\),

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

On the other hand, by the definition of \(u^t=\Hom(u,A)\) we have \(u^t(\xi)=\xi\circ u\). Hence the above equation can be written as

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

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

Moreover, for two \(A\)-linear maps \(u,v:M \rightarrow N\) and arbitrary \(\xi\in N^\ast\), \(x\in M\), the equality \(\xi((u+v)(x))=\xi(u(x))+\xi(v(x))\), together with the functoriality of \(\Hom(-,A)\) and [Algebraic Structures] §Modules, ⁋Proposition 8, yields the following proposition.

Proposition 5 The following hold.

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

Dual Basis

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

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

Then, taking the dual of this isomorphism, we obtain an isomorphism of right \(A\)-modules

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

Now among the elements of the right-hand side, consider those whose \(i\)-th component is \(1\) and all other components are \(0\), and write \(e_i^\ast\) for the preimage of such an element under \(\varepsilon^t\). Then we know that

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

holds. The collection of these elements is linearly independent, but if \(I\) is infinite then 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 these elements do form 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 \((e_i)\).

Double Dual Space

For any left \(A\)-module \(M\), the dual \(M^\ast\) is a right \(A\)-module, and the dual \(M^{\ast\ast}\) of \(M^\ast\) is again a left \(A\)-module. On the other hand, for arbitrary \(x\in M\), one can check that the function

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

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

Definition 7 If the above function \(M \rightarrow M^{\ast\ast}\) is bijective, we call \(M\) 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.

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References

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