Deforming metrics from non-negative to positive Ricci curvature












9












$begingroup$


Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dimgeq 3$, when can we deform the metric to a positive Ricci curved one?



I know it's impossible in general due to the flat factor in the universal covering. But what about we add some topological restrictions on $M$ like simply connectedness? Are there any positive or negative results on this problem?



( Besides, are there now any examples of simply connected closed manifold with positive scalar curved metric by do not admit a positive Ricci curved metric? )



------------------------------------------------------------update 1------------------------------------------------------------



Thanks to the answer by Robert, I may simplify the problem in the following sense,



Given a simply connected flat Einstein manifold $(M,g)$ with $hat{A}$-genus non-vanishing, and set $(mathbb{S}^2,h)$ be the standard unit sphere, can $(Mtimes mathbb{S}^2, g+h)$ be perturbed to a Ricci-positive manifold? $ $In general, what about changing $(mathbb{S}^2,h)$ to an arbitrary closed Ricci-positive manifold?






dg.differential-geometry riemannian-geometry ricci-curvature







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is better to ask the follow up question separately.
    $endgroup$
    – Igor Belegradek
    Nov 29 '18 at 22:36










  • $begingroup$
    On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
    $endgroup$
    – Igor Belegradek
    Nov 30 '18 at 1:35
















9












$begingroup$


Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dimgeq 3$, when can we deform the metric to a positive Ricci curved one?



I know it's impossible in general due to the flat factor in the universal covering. But what about we add some topological restrictions on $M$ like simply connectedness? Are there any positive or negative results on this problem?



( Besides, are there now any examples of simply connected closed manifold with positive scalar curved metric by do not admit a positive Ricci curved metric? )



------------------------------------------------------------update 1------------------------------------------------------------



Thanks to the answer by Robert, I may simplify the problem in the following sense,



Given a simply connected flat Einstein manifold $(M,g)$ with $hat{A}$-genus non-vanishing, and set $(mathbb{S}^2,h)$ be the standard unit sphere, can $(Mtimes mathbb{S}^2, g+h)$ be perturbed to a Ricci-positive manifold? $ $In general, what about changing $(mathbb{S}^2,h)$ to an arbitrary closed Ricci-positive manifold?






dg.differential-geometry riemannian-geometry ricci-curvature







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is better to ask the follow up question separately.
    $endgroup$
    – Igor Belegradek
    Nov 29 '18 at 22:36










  • $begingroup$
    On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
    $endgroup$
    – Igor Belegradek
    Nov 30 '18 at 1:35














9












9








9


1



$begingroup$


Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dimgeq 3$, when can we deform the metric to a positive Ricci curved one?



I know it's impossible in general due to the flat factor in the universal covering. But what about we add some topological restrictions on $M$ like simply connectedness? Are there any positive or negative results on this problem?



( Besides, are there now any examples of simply connected closed manifold with positive scalar curved metric by do not admit a positive Ricci curved metric? )



------------------------------------------------------------update 1------------------------------------------------------------



Thanks to the answer by Robert, I may simplify the problem in the following sense,



Given a simply connected flat Einstein manifold $(M,g)$ with $hat{A}$-genus non-vanishing, and set $(mathbb{S}^2,h)$ be the standard unit sphere, can $(Mtimes mathbb{S}^2, g+h)$ be perturbed to a Ricci-positive manifold? $ $In general, what about changing $(mathbb{S}^2,h)$ to an arbitrary closed Ricci-positive manifold?






dg.differential-geometry riemannian-geometry ricci-curvature







share|cite|improve this question











$endgroup$




Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dimgeq 3$, when can we deform the metric to a positive Ricci curved one?



I know it's impossible in general due to the flat factor in the universal covering. But what about we add some topological restrictions on $M$ like simply connectedness? Are there any positive or negative results on this problem?



( Besides, are there now any examples of simply connected closed manifold with positive scalar curved metric by do not admit a positive Ricci curved metric? )



------------------------------------------------------------update 1------------------------------------------------------------



Thanks to the answer by Robert, I may simplify the problem in the following sense,



Given a simply connected flat Einstein manifold $(M,g)$ with $hat{A}$-genus non-vanishing, and set $(mathbb{S}^2,h)$ be the standard unit sphere, can $(Mtimes mathbb{S}^2, g+h)$ be perturbed to a Ricci-positive manifold? $ $In general, what about changing $(mathbb{S}^2,h)$ to an arbitrary closed Ricci-positive manifold?






dg.differential-geometry riemannian-geometry ricci-curvature




dg.differential-geometry riemannian-geometry ricci-curvature






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 21:19







ZHans Wang

















asked Nov 26 '18 at 4:06









ZHans WangZHans Wang

484




484












  • $begingroup$
    It is better to ask the follow up question separately.
    $endgroup$
    – Igor Belegradek
    Nov 29 '18 at 22:36










  • $begingroup$
    On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
    $endgroup$
    – Igor Belegradek
    Nov 30 '18 at 1:35


















  • $begingroup$
    It is better to ask the follow up question separately.
    $endgroup$
    – Igor Belegradek
    Nov 29 '18 at 22:36










  • $begingroup$
    On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
    $endgroup$
    – Igor Belegradek
    Nov 30 '18 at 1:35
















$begingroup$
It is better to ask the follow up question separately.
$endgroup$
– Igor Belegradek
Nov 29 '18 at 22:36




$begingroup$
It is better to ask the follow up question separately.
$endgroup$
– Igor Belegradek
Nov 29 '18 at 22:36












$begingroup$
On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
$endgroup$
– Igor Belegradek
Nov 30 '18 at 1:35




$begingroup$
On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
$endgroup$
– Igor Belegradek
Nov 30 '18 at 1:35










2 Answers
2






active

oldest

votes


















9












$begingroup$

There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $hat A$-genus must vanish.



If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $mathrm{G}_2$ (in dimension $7$), $mathrm{Spin}(7)$ (in dimension $8$), or holonomy in $mathrm{SU}(n)$ (in dimension $2n$) whose $hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
    $endgroup$
    – ZHans Wang
    Nov 27 '18 at 5:03



















8












$begingroup$

This is not a complete answer but would be helpful. Here are a few facts:



Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is
non-negative and positive somewhere, then the manifold carries a metric with
positive Ricci curvature.



Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly
positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).



Relation with scalar curvature:




  1. There are still no known examples of simply connected manifolds that admit
    positive scalar curvature but not positive Ricci curvature


  2. If a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.



This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "504"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f316209%2fdeforming-metrics-from-non-negative-to-positive-ricci-curvature%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $hat A$-genus must vanish.



    If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $mathrm{G}_2$ (in dimension $7$), $mathrm{Spin}(7)$ (in dimension $8$), or holonomy in $mathrm{SU}(n)$ (in dimension $2n$) whose $hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
      $endgroup$
      – ZHans Wang
      Nov 27 '18 at 5:03
















    9












    $begingroup$

    There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $hat A$-genus must vanish.



    If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $mathrm{G}_2$ (in dimension $7$), $mathrm{Spin}(7)$ (in dimension $8$), or holonomy in $mathrm{SU}(n)$ (in dimension $2n$) whose $hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
      $endgroup$
      – ZHans Wang
      Nov 27 '18 at 5:03














    9












    9








    9





    $begingroup$

    There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $hat A$-genus must vanish.



    If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $mathrm{G}_2$ (in dimension $7$), $mathrm{Spin}(7)$ (in dimension $8$), or holonomy in $mathrm{SU}(n)$ (in dimension $2n$) whose $hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.






    share|cite|improve this answer









    $endgroup$



    There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $hat A$-genus must vanish.



    If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $mathrm{G}_2$ (in dimension $7$), $mathrm{Spin}(7)$ (in dimension $8$), or holonomy in $mathrm{SU}(n)$ (in dimension $2n$) whose $hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 26 '18 at 12:33









    Robert BryantRobert Bryant

    73.9k6216319




    73.9k6216319












    • $begingroup$
      This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
      $endgroup$
      – ZHans Wang
      Nov 27 '18 at 5:03


















    • $begingroup$
      This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
      $endgroup$
      – ZHans Wang
      Nov 27 '18 at 5:03
















    $begingroup$
    This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
    $endgroup$
    – ZHans Wang
    Nov 27 '18 at 5:03




    $begingroup$
    This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
    $endgroup$
    – ZHans Wang
    Nov 27 '18 at 5:03











    8












    $begingroup$

    This is not a complete answer but would be helpful. Here are a few facts:



    Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is
    non-negative and positive somewhere, then the manifold carries a metric with
    positive Ricci curvature.



    Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly
    positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).



    Relation with scalar curvature:




    1. There are still no known examples of simply connected manifolds that admit
      positive scalar curvature but not positive Ricci curvature


    2. If a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.



    This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$

      This is not a complete answer but would be helpful. Here are a few facts:



      Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is
      non-negative and positive somewhere, then the manifold carries a metric with
      positive Ricci curvature.



      Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly
      positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).



      Relation with scalar curvature:




      1. There are still no known examples of simply connected manifolds that admit
        positive scalar curvature but not positive Ricci curvature


      2. If a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.



      This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        This is not a complete answer but would be helpful. Here are a few facts:



        Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is
        non-negative and positive somewhere, then the manifold carries a metric with
        positive Ricci curvature.



        Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly
        positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).



        Relation with scalar curvature:




        1. There are still no known examples of simply connected manifolds that admit
          positive scalar curvature but not positive Ricci curvature


        2. If a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.



        This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf






        share|cite|improve this answer









        $endgroup$



        This is not a complete answer but would be helpful. Here are a few facts:



        Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is
        non-negative and positive somewhere, then the manifold carries a metric with
        positive Ricci curvature.



        Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly
        positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).



        Relation with scalar curvature:




        1. There are still no known examples of simply connected manifolds that admit
          positive scalar curvature but not positive Ricci curvature


        2. If a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.



        This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 '18 at 6:33









        C.F.GC.F.G

        1,27821035




        1,27821035






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to MathOverflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f316209%2fdeforming-metrics-from-non-negative-to-positive-ricci-curvature%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Costa Masnaga

            Image
            .mw-parser-output .avviso .mbox-text-div>div,.mw-parser-output .avviso .mbox-text-full-div>div{font-size:90%}.mw-parser-output .avviso .mbox-image div{width:52px}.mw-parser-output .avviso .mbox-text-full-div .hide-when-compact{display:block} Questa voce o sezione sull'argomento Lombardia non cita le fonti necessarie o quelle presenti sono insufficienti . Puoi migliorare questa voce aggiungendo citazioni da fonti attendibili secondo le linee guida sull'uso delle fonti. Segui i suggerimenti del progetto di riferimento. Costa Masnaga comune Localizzazione Stato   Italia Regione Lombardia Provincia Lecco Amministrazione Sindaco Sabina Panzeri (Al centro il cittadino) dall'08/06/2009 Territorio Coordinate 45°46′N 9°17′E  /  45.766667°N 9.283333°E 45.766667; 9.283333  ( Costa Masnaga ) Coordinate: 45°46′N 9°17′E  /  45.766667°N 9.283333°E 45.766667; 9.283333  ( Costa Masnaga ) Altitudine 318 m s.l.m...
            Read more

            Fotorealismo

            Image
            Guerrino Boatto, Venice's St. Barnaba Church and canal Il fotorealismo è un genere di pittura basato sull'uso di una o più fotografie dalle quali l'artista prende le informazioni necessarie da utilizzare poi nel processo della creazione dell'opera, in genere ad olio o ad acrilico, al fine di avere un aspetto finale il più simile possibile alla fotografia. Indice 1 Storia 2 Lista dei fotorealisti 3 Note 4 Voci correlate 5 Altri progetti Storia | Il fotorealismo è un movimento che nasce negli Stati Uniti d'America alla fine degli anni sessanta [1] sulla scia della Pop Art [2] [3] [4] , e insieme condividono il carattere reazionario verso la travolgente ascesa dei media, che alla metà del ventesimo secolo si trasforma in un fenomeno di massa talmente imponente da minacciare il valore dell'immagine nell'arte. [5] [6] Ne consegue così che, colto il gesto d'inizio che fu della Pop Art, i fotorealisti aprono ad un ...
            Read more

            Sidney Franklin

            Image
            Questa voce sull'argomento registi statunitensi è solo un abbozzo . Contribuisci a migliorarla secondo le convenzioni di Wikipedia. Sidney Franklin, foto di Carl Van Vechten Oscar alla memoria Irving G. Thalberg 1943 Sidney Franklin , all'anagrafe Sidney Arnold Franklin , (San Francisco, 21 marzo 1893 – Santa Monica, 18 maggio 1972), è stato un regista, produttore cinematografico e attore statunitense. Il 20 giugno 1963, si sposò con l'attrice australiana Enid Bennett. Il matrimonio durò fino alla morte dell'attrice, il 14 maggio 1969 [1] Indice 1 Filmografia 1.1 Regista 1.2 Aiuto regia (parziale) 1.3 Produttore 1.4 Attore (parziale) 2 Note 3 Altri progetti 4 Collegamenti esterni Filmografia | Regista | The Baby , co-regia Chester M. Franklin (1915) The Rivals , co-regia Chester M. Franklin (1915) Little Dick's First Case , co-regia Chester M. Franklin (1915) Her Filmland Hero , co-...
            Read more