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    • CommentRowNumber1.
    • CommentAuthorDmitri Pavlov
    • CommentTimeApr 14th 2021

    Created with the following content:



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

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

    Furthermore, the abstract evaluation map

    Hom R(M,R) RMRHom_R(M,R)\otimes_R M\to R

    must coincide with the map induced by the bilinear map

    Hom R(M,R)× RMRHom_R(M,R)\times_R M\to R

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


    \begin{theorem} An RR-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 nR^n for some n0n\ge0, so to show that finitely generated projective modules are dualizable, it suffices to observe that RR is dualizable as an RR-module.

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

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

    and the evaluation map

    ev:Hom R(M,R)MRev: Hom_R(M,R)\otimes M\to R

    satisfy the triangle identities:

    (id Mev)(coevid M)=id M,(id_M \otimes ev)\circ (coev\otimes id_M) = id_M, (evid Hom(M,R))(id Hom(M,R)coev)=id Hom(M,R).(ev \otimes id_{Hom(M,R)})\circ (id_{Hom(M,R)}\otimes coev) = id_{Hom(M,R)}.

    The coevaluation map sends 1R1\in R to a finite sum

    iIm if i.\sum_{i\in I} m_i\otimes f_i.

    The triangle identities now read

    iIm if i(p)=p,pM\sum_{i\in I} m_i f_i(p) = p,\qquad p\in M iIr(m i)f i=r,rHom R(M,R).\sum_{i\in I} r(m_i) f_i = r, \qquad r\in Hom_R(M,R).

    The first identity implies that m im_i (iIi\in I) generate MM as an RR-module, i.e., MM is finitely generated.

    Consider the map a:R IMa: R^I\to M that sends (r i) iI(r_i)_{i\in I} to iIm ir i\sum_{i\in I} m_i r_i. Consider also the map b:MR Ib: M\to R^I that sends pMp\in M to (f i(p)) iIR I(f_i(p))_{i\in I}\in R^I. The first triangle identity now reads ba=id Mb a = id_M. Thus, MM is a retract of R IR^I, i.e., MM is a projective module. \end{proof}

    Related concepts

    v1, current

    • CommentRowNumber2.
    • CommentAuthorDmitri Pavlov
    • CommentTimeApr 14th 2021

    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 RR 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 RR. \end{theorem}

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

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

    v1, current

  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 im if i\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.

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