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1. • CommentRowNumber1.
   • CommentAuthorDmitri Pavlov
   • CommentTimeApr 14th 2021
   • PermaLink
   Author: Dmitri Pavlov
   Format: MarkdownItexCreated with the following content: ## Idea ## Definition A [[module]] $M$ over a [[commutative ring]] $R$ is __dualizable__ if it is a [[dualizable object]] in the [[symmetric monoidal category]] of $R$-[[modules]] equipped with the [[tensor product]] over $R$. Since this [[[symmetric monoidal category]] is a [[closed monoidal category]], the [[dual object]] to $M$ is necessarily $Hom_R(M,R)$. Furthermore, the abstract [[evaluation map]] $$Hom_R(M,R)\otimes_R M\to R$$ must coincide with the map induced by the [[bilinear map]] $$Hom_R(M,R)\times_R M\to R$$ that sends $(f,m)$ to $f(m)$. ## Characterization \begin{theorem} An $R$-module is dualizable if and only if it is a [[finitely generated]] [[projective module]]. \end{theorem} \begin{proof} First, [[dualizable objects]] are closed under [[retracts]] and finite [[direct sums]]. Any [[finitely generated]] [[projective module]] is a [[retract]] of $R^n$ for some $n\ge0$, so to show that [[finitely generated]] [[projective modules]] are [[dualizable]], it suffices to observe that $R$ is dualizable as an $R$-[[module]]. Conversely, we show that [[dualizable objects]] are [[finitely generated]] [[projective modules]]. Unfolding the definition of a [[dualizable object]], an $R$-[[module]] $M$ is dualizable if the coevaluation map $$coev: R \to M\otimes Hom_R(M,R)$$ and the evaluation map $$ev: Hom_R(M,R)\otimes M\to R$$ satisfy the [[triangle identities]]: $$(id_M \otimes ev)\circ (coev\otimes id_M) = id_M,$$ $$(ev \otimes id_{Hom(M,R)})\circ (id_{Hom(M,R)}\otimes coev) = id_{Hom(M,R)}.$$ The coevaluation map sends $1\in R$ to a _finite_ sum $$\sum_{i\in I} m_i\otimes f_i.$$ The [[triangle identities]] now read $$\sum_{i\in I} m_i f_i(p) = p,\qquad p\in M$$ $$\sum_{i\in I} r(m_i) f_i = r, \qquad r\in Hom_R(M,R).$$ The first identity implies that $m_i$ ($i\in I$) generate $M$ as an $R$-module, i.e., $M$ is [[finitely generated]]. Consider the map $a: R^I\to M$ that sends $(r_i)_{i\in I}$ to $\sum_{i\in I} m_i r_i$. Consider also the map $b: M\to R^I$ that sends $p\in M$ to $(f_i(p))_{i\in I}\in R^I$. The first triangle identity now reads $b a = id_M$. Thus, $M$ is a retract of $R^I$, i.e., $M$ is a [[projective module]]. \end{proof} ## Related concepts * [[dualizable object]] * [[perfect chain complex]] * [[perfect module]] <a href="https://ncatlab.org/nlab/revision/dualizable+module/1">v1</a>, <a href="https://ncatlab.org/nlab/show/dualizable+module">current</a>

   Created with the following content:

   Idea

   Definition

   A module $M$ over a commutative ring $R$ is dualizable if it is a dualizable object in the symmetric monoidal category of $R$-modules equipped with the tensor product over $R$.

   Since this [symmetric monoidal category is a closed monoidal category, the dual object to $M$ is necessarily $Hom_R(M,R)$.

   Furthermore, the abstract evaluation map

   $$Hom_R(M,R)\otimes_R M\to R$$

   must coincide with the map induced by the bilinear map

   $$Hom_R(M,R)\times_R M\to R$$

   that sends $(f,m)$ to $f(m)$.

   Characterization

   \begin{theorem} An $R$-module is dualizable if and only if it is a finitely generated projective module. \end{theorem}

   \begin{proof} First, dualizable objects are closed under retracts and finite direct sums. Any finitely generated projective module is a retract of $R^n$ for some $n\ge0$, so to show that finitely generated projective modules are dualizable, it suffices to observe that $R$ is dualizable as an $R$-module.

   Conversely, we show that dualizable objects are finitely generated projective modules. Unfolding the definition of a dualizable object, an $R$-module $M$ is dualizable if the coevaluation map

   $$coev: R \to M\otimes Hom_R(M,R)$$

   and the evaluation map

   $$ev: Hom_R(M,R)\otimes M\to R$$

   satisfy the triangle identities:

   $$(id_M \otimes ev)\circ (coev\otimes id_M) = id_M,$$ $$(ev \otimes id_{Hom(M,R)})\circ (id_{Hom(M,R)}\otimes coev) = id_{Hom(M,R)}.$$

   The coevaluation map sends $1\in R$ to a finite sum

   $$\sum_{i\in I} m_i\otimes f_i.$$

   The triangle identities now read

   $$\sum_{i\in I} m_i f_i(p) = p,\qquad p\in M$$ $$\sum_{i\in I} r(m_i) f_i = r, \qquad r\in Hom_R(M,R).$$

   The first identity implies that $m_i$ ($i\in I$) generate $M$ as an $R$-module, i.e., $M$ is finitely generated.

   Consider the map $a: R^I\to M$ that sends $(r_i)_{i\in I}$ to $\sum_{i\in I} m_i r_i$. Consider also the map $b: M\to R^I$ that sends $p\in M$ to $(f_i(p))_{i\in I}\in R^I$. The first triangle identity now reads $b a = id_M$. Thus, $M$ is a retract of $R^I$, i.e., $M$ is a projective module. \end{proof}

   Related concepts

   • dualizable object

   • perfect chain complex

   • perfect module

   v1, current

2. • CommentRowNumber2.
   • CommentAuthorDmitri Pavlov
   • CommentTimeApr 14th 2021
   • PermaLink
   Author: Dmitri Pavlov
   Format: MarkdownItexAdded this: ## Geometric interpretation See also [[Serre–Swan theorem]] and [[smooth Serre–Swan theorem]]. \begin{theorem} (Serre, 1955.) The [[category]] of dualizable modules over a [[commutative ring]] $R$ is equivalent to the [[category]] of [[algebraic vector bundles]] (defined as [[locally free sheaves]] over the [[structure sheaf of rings]]) over the [[Zariski spectrum]] of $R$. \end{theorem} \begin{theorem} (Swan, 1962.) Given a [[compact Hausdorff space]] $X$, the [[category]] of dualizable modules over the real [[algebra]] of [[continuous maps]] $X\to\mathbf{R}$ is equivalent to the category of finite-dimensional continuous [[vector bundles]] over $X$. \end{theorem} \begin{theorem} (See, e.g., Nestruev 2003, 11.33.) Given a [[smooth manifold]] $X$, the [[category]] of dualizable modules over the real [[algebra]] of [[smooth maps]] $X\to\mathbf{R}$ is equivalent to the category of finite-dimensional smooth [[vector bundles]] over $X$. \end{theorem} <a href="https://ncatlab.org/nlab/revision/dualizable+module/1">v1</a>, <a href="https://ncatlab.org/nlab/show/dualizable+module">current</a>

   Added this:

   Geometric interpretation

   See also Serre–Swan theorem and smooth Serre–Swan theorem.

   \begin{theorem} (Serre, 1955.) The category of dualizable modules over a commutative ring $R$ is equivalent to the category of algebraic vector bundles (defined as locally free sheaves over the structure sheaf of rings) over the Zariski spectrum of $R$. \end{theorem}

   \begin{theorem} (Swan, 1962.) Given a compact Hausdorff space $X$, the category of dualizable modules over the real algebra of continuous maps $X\to\mathbf{R}$ is equivalent to the category of finite-dimensional continuous vector bundles over $X$. \end{theorem}

   \begin{theorem} (See, e.g., Nestruev 2003, 11.33.) Given a smooth manifold $X$, the category of dualizable modules over the real algebra of smooth maps $X\to\mathbf{R}$ is equivalent to the category of finite-dimensional smooth vector bundles over $X$. \end{theorem}

   v1, current

3. • CommentRowNumber3.
   • CommentAuthorOscar_Cunningham
   • CommentTimeApr 20th 2021
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   Author: Oscar_Cunningham
   Format: MarkdownItexHow constructive is the equivalence between dualizable and finitely generated projective? It feels a lot like choosing a basis. In particular the step where we get a specific sum $\sum_i m_i\otimes f_i$ seems like we're getting something for nothing.

   How constructive is the equivalence between dualizable and finitely generated projective? It feels a lot like choosing a basis. In particular the step where we get a specific sum $\sum_i m_i\otimes f_i$ seems like we’re getting something for nothing.

4. • CommentRowNumber4.
   • CommentAuthorDmitri Pavlov
   • CommentTimeApr 20th 2021
   • PermaLink
   Author: Dmitri Pavlov
   Format: MarkdownItexRe #3: Looks constructive to me. Concerning the specific sum: the tensor product of M and N over A (also constructively, I presume) is the quotient of the free vector space on M⨯N by the vector space generated by the elements (m+m',n)-(m,n)-(m',n), (am,n)-a(m,n), and the symmetric versions. So we get a finite sum of m_i⊗f_i because it is a representative of an equivalence class in the quotient.

   Re #3: Looks constructive to me.

   Concerning the specific sum: the tensor product of M and N over A (also constructively, I presume) is the quotient of the free vector space on M⨯N by the vector space generated by the elements (m+m’,n)-(m,n)-(m’,n), (am,n)-a(m,n), and the symmetric versions.

   So we get a finite sum of m_i⊗f_i because it is a representative of an equivalence class in the quotient.