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• CommentRowNumber1.
• CommentAuthorDmitri Pavlov
• CommentTimeApr 14th 2021

Created with the following content:

## 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

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

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

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

• CommentRowNumber2.
• CommentAuthorDmitri Pavlov
• CommentTimeApr 14th 2021

## Geometric interpretation

\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}

1. 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.

• CommentRowNumber4.
• CommentAuthorDmitri Pavlov
• CommentTimeApr 20th 2021

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.