induced connection on open sets












3












$begingroup$


Let $M$ be a smooth manifold. If $nabla$ is a linear connection on $M$, I would like to induce a unique linear connection on an open subset $Usubseteq M$. I know that for all $pin U$ there is a natural isomorphism $T_pUcong T_pM$, so I can restrict global vector fields to local vector fields on $U$. Unfortunately there are some local vector fields on $U$ that don't came from a restriction of global vector fields.



For this reason I can't find a reasonable linear connection $nabla^U$ over $U$ induced by $nabla$. I need help.










share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Let $M$ be a smooth manifold. If $nabla$ is a linear connection on $M$, I would like to induce a unique linear connection on an open subset $Usubseteq M$. I know that for all $pin U$ there is a natural isomorphism $T_pUcong T_pM$, so I can restrict global vector fields to local vector fields on $U$. Unfortunately there are some local vector fields on $U$ that don't came from a restriction of global vector fields.



    For this reason I can't find a reasonable linear connection $nabla^U$ over $U$ induced by $nabla$. I need help.










    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let $M$ be a smooth manifold. If $nabla$ is a linear connection on $M$, I would like to induce a unique linear connection on an open subset $Usubseteq M$. I know that for all $pin U$ there is a natural isomorphism $T_pUcong T_pM$, so I can restrict global vector fields to local vector fields on $U$. Unfortunately there are some local vector fields on $U$ that don't came from a restriction of global vector fields.



      For this reason I can't find a reasonable linear connection $nabla^U$ over $U$ induced by $nabla$. I need help.










      share|cite|improve this question









      $endgroup$




      Let $M$ be a smooth manifold. If $nabla$ is a linear connection on $M$, I would like to induce a unique linear connection on an open subset $Usubseteq M$. I know that for all $pin U$ there is a natural isomorphism $T_pUcong T_pM$, so I can restrict global vector fields to local vector fields on $U$. Unfortunately there are some local vector fields on $U$ that don't came from a restriction of global vector fields.



      For this reason I can't find a reasonable linear connection $nabla^U$ over $U$ induced by $nabla$. I need help.







      differential-geometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 8 '12 at 17:19









      DubiousDubious

      3,36472675




      3,36472675






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          The connection $nabla$ on a manifold $M$ is a local operator. The value of $nabla_X(Y)$ at a point $p in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
            $endgroup$
            – Dubious
            Dec 8 '12 at 17:58





















          2












          $begingroup$

          Take an open cover ${U_j}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity ${phi_j}$ subordinated to ${U_j}$ (i.e. $mathrm{supp}phi_jsubset U_j$).



          For any vector-field $X$ on $U$, $X=sum phi_jcdot X$ and $phi_j X$ is a vector-field on $M$. Therefore we can define
          $$nabla^U_XY=sumnabla_{phi_j X}Y$$
          and
          $$nabla^U_YX=sumnabla_Yphi_j X$$
          The definition is locally meaningful, because the covering is locally finite, and so are the sums.



          Hope it helps.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
            $endgroup$
            – Dubious
            Dec 8 '12 at 17:42






          • 1




            $begingroup$
            The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
            $endgroup$
            – Jason DeVito
            Dec 8 '12 at 18:01








          • 1




            $begingroup$
            @Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
            $endgroup$
            – Jason DeVito
            Dec 8 '12 at 18:16








          • 1




            $begingroup$
            No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
            $endgroup$
            – wisefool
            Dec 8 '12 at 18:18






          • 1




            $begingroup$
            Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
            $endgroup$
            – wisefool
            Dec 8 '12 at 18:21



















          0












          $begingroup$

          For a connection $nabla$ on a vector bundle $E$ over $M$ one can show two things:




          1. If two sections $X,Y$ of $TM$ agree at a point $p$ then $(nabla_Xpsi)_p=(nabla_Ypsi)_p$ for all sections $psi$ of $E$.


          2. If two sections $phi,psi$ agree locally arround $p$ then $(nabla_Xphi)_p=(nabla_Xpsi)_p$ for all sections $X$ of $TM$.



          Furthermore:




          1. Any $X_pin T_pM$ can be extended to a section of $TM$.


          2. For any open neighbourhood $U$ of $p$ and any $psiin E_{|U}$ there exists a section $tilde{psi}$ of $E$ which locally arround $p$ agrees with $psi$ (just take a chart multiply by a suitable bump function and set it $0$ elsewhere).



          Now you can define the induced connection $nabla'$ as follows:



          $$(nabla'_Xpsi)_p=(nabla_tilde{X}tilde psi)_p $$



          where $tilde{X}$ is any extension of $X_p$ and $tilde psi$ is any local extension of $psi$. By the above this is independent of the choice of $tilde{X}$ and $tilde psi$.






          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: "69"
            };
            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%2fmath.stackexchange.com%2fquestions%2f253880%2finduced-connection-on-open-sets%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            The connection $nabla$ on a manifold $M$ is a local operator. The value of $nabla_X(Y)$ at a point $p in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
              $endgroup$
              – Dubious
              Dec 8 '12 at 17:58


















            2












            $begingroup$

            The connection $nabla$ on a manifold $M$ is a local operator. The value of $nabla_X(Y)$ at a point $p in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
              $endgroup$
              – Dubious
              Dec 8 '12 at 17:58
















            2












            2








            2





            $begingroup$

            The connection $nabla$ on a manifold $M$ is a local operator. The value of $nabla_X(Y)$ at a point $p in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.






            share|cite|improve this answer









            $endgroup$



            The connection $nabla$ on a manifold $M$ is a local operator. The value of $nabla_X(Y)$ at a point $p in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 8 '12 at 17:32









            levaplevap

            47.9k33274




            47.9k33274












            • $begingroup$
              The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
              $endgroup$
              – Dubious
              Dec 8 '12 at 17:58




















            • $begingroup$
              The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
              $endgroup$
              – Dubious
              Dec 8 '12 at 17:58


















            $begingroup$
            The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
            $endgroup$
            – Dubious
            Dec 8 '12 at 17:58






            $begingroup$
            The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $nabla_XY=nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem.
            $endgroup$
            – Dubious
            Dec 8 '12 at 17:58













            2












            $begingroup$

            Take an open cover ${U_j}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity ${phi_j}$ subordinated to ${U_j}$ (i.e. $mathrm{supp}phi_jsubset U_j$).



            For any vector-field $X$ on $U$, $X=sum phi_jcdot X$ and $phi_j X$ is a vector-field on $M$. Therefore we can define
            $$nabla^U_XY=sumnabla_{phi_j X}Y$$
            and
            $$nabla^U_YX=sumnabla_Yphi_j X$$
            The definition is locally meaningful, because the covering is locally finite, and so are the sums.



            Hope it helps.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
              $endgroup$
              – Dubious
              Dec 8 '12 at 17:42






            • 1




              $begingroup$
              The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
              $endgroup$
              – Jason DeVito
              Dec 8 '12 at 18:01








            • 1




              $begingroup$
              @Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
              $endgroup$
              – Jason DeVito
              Dec 8 '12 at 18:16








            • 1




              $begingroup$
              No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
              $endgroup$
              – wisefool
              Dec 8 '12 at 18:18






            • 1




              $begingroup$
              Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
              $endgroup$
              – wisefool
              Dec 8 '12 at 18:21
















            2












            $begingroup$

            Take an open cover ${U_j}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity ${phi_j}$ subordinated to ${U_j}$ (i.e. $mathrm{supp}phi_jsubset U_j$).



            For any vector-field $X$ on $U$, $X=sum phi_jcdot X$ and $phi_j X$ is a vector-field on $M$. Therefore we can define
            $$nabla^U_XY=sumnabla_{phi_j X}Y$$
            and
            $$nabla^U_YX=sumnabla_Yphi_j X$$
            The definition is locally meaningful, because the covering is locally finite, and so are the sums.



            Hope it helps.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
              $endgroup$
              – Dubious
              Dec 8 '12 at 17:42






            • 1




              $begingroup$
              The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
              $endgroup$
              – Jason DeVito
              Dec 8 '12 at 18:01








            • 1




              $begingroup$
              @Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
              $endgroup$
              – Jason DeVito
              Dec 8 '12 at 18:16








            • 1




              $begingroup$
              No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
              $endgroup$
              – wisefool
              Dec 8 '12 at 18:18






            • 1




              $begingroup$
              Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
              $endgroup$
              – wisefool
              Dec 8 '12 at 18:21














            2












            2








            2





            $begingroup$

            Take an open cover ${U_j}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity ${phi_j}$ subordinated to ${U_j}$ (i.e. $mathrm{supp}phi_jsubset U_j$).



            For any vector-field $X$ on $U$, $X=sum phi_jcdot X$ and $phi_j X$ is a vector-field on $M$. Therefore we can define
            $$nabla^U_XY=sumnabla_{phi_j X}Y$$
            and
            $$nabla^U_YX=sumnabla_Yphi_j X$$
            The definition is locally meaningful, because the covering is locally finite, and so are the sums.



            Hope it helps.






            share|cite|improve this answer









            $endgroup$



            Take an open cover ${U_j}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity ${phi_j}$ subordinated to ${U_j}$ (i.e. $mathrm{supp}phi_jsubset U_j$).



            For any vector-field $X$ on $U$, $X=sum phi_jcdot X$ and $phi_j X$ is a vector-field on $M$. Therefore we can define
            $$nabla^U_XY=sumnabla_{phi_j X}Y$$
            and
            $$nabla^U_YX=sumnabla_Yphi_j X$$
            The definition is locally meaningful, because the covering is locally finite, and so are the sums.



            Hope it helps.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 8 '12 at 17:33









            wisefoolwisefool

            3,179711




            3,179711












            • $begingroup$
              why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
              $endgroup$
              – Dubious
              Dec 8 '12 at 17:42






            • 1




              $begingroup$
              The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
              $endgroup$
              – Jason DeVito
              Dec 8 '12 at 18:01








            • 1




              $begingroup$
              @Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
              $endgroup$
              – Jason DeVito
              Dec 8 '12 at 18:16








            • 1




              $begingroup$
              No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
              $endgroup$
              – wisefool
              Dec 8 '12 at 18:18






            • 1




              $begingroup$
              Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
              $endgroup$
              – wisefool
              Dec 8 '12 at 18:21


















            • $begingroup$
              why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
              $endgroup$
              – Dubious
              Dec 8 '12 at 17:42






            • 1




              $begingroup$
              The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
              $endgroup$
              – Jason DeVito
              Dec 8 '12 at 18:01








            • 1




              $begingroup$
              @Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
              $endgroup$
              – Jason DeVito
              Dec 8 '12 at 18:16








            • 1




              $begingroup$
              No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
              $endgroup$
              – wisefool
              Dec 8 '12 at 18:18






            • 1




              $begingroup$
              Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
              $endgroup$
              – wisefool
              Dec 8 '12 at 18:21
















            $begingroup$
            why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
            $endgroup$
            – Dubious
            Dec 8 '12 at 17:42




            $begingroup$
            why $phi_jX$ is a vector field on $M$? For every $pin U$ we have that $(phi_jX)_p=phi(p)X_p$, but if $p$ is in $Xsetminus U$ what is the sense?
            $endgroup$
            – Dubious
            Dec 8 '12 at 17:42




            1




            1




            $begingroup$
            The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
            $endgroup$
            – Jason DeVito
            Dec 8 '12 at 18:01






            $begingroup$
            The point is that on $Xsetminus U$, we have $phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $phi(p) X_p = 0_p$. So $phi X$ is still a smooth vector field on all of $M$.
            $endgroup$
            – Jason DeVito
            Dec 8 '12 at 18:01






            1




            1




            $begingroup$
            @Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
            $endgroup$
            – Jason DeVito
            Dec 8 '12 at 18:16






            $begingroup$
            @Galoisfan: Well, I'm thinking more about $phi X$ than $sum phi_j X$. Note that $phi X$ is not an extension of $X$ to all of $M$ because there are points $pin U$ with $phi(p)X_p neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$.
            $endgroup$
            – Jason DeVito
            Dec 8 '12 at 18:16






            1




            1




            $begingroup$
            No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
            $endgroup$
            – wisefool
            Dec 8 '12 at 18:18




            $begingroup$
            No, $sumphi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates.
            $endgroup$
            – wisefool
            Dec 8 '12 at 18:18




            1




            1




            $begingroup$
            Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
            $endgroup$
            – wisefool
            Dec 8 '12 at 18:21




            $begingroup$
            Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works.
            $endgroup$
            – wisefool
            Dec 8 '12 at 18:21











            0












            $begingroup$

            For a connection $nabla$ on a vector bundle $E$ over $M$ one can show two things:




            1. If two sections $X,Y$ of $TM$ agree at a point $p$ then $(nabla_Xpsi)_p=(nabla_Ypsi)_p$ for all sections $psi$ of $E$.


            2. If two sections $phi,psi$ agree locally arround $p$ then $(nabla_Xphi)_p=(nabla_Xpsi)_p$ for all sections $X$ of $TM$.



            Furthermore:




            1. Any $X_pin T_pM$ can be extended to a section of $TM$.


            2. For any open neighbourhood $U$ of $p$ and any $psiin E_{|U}$ there exists a section $tilde{psi}$ of $E$ which locally arround $p$ agrees with $psi$ (just take a chart multiply by a suitable bump function and set it $0$ elsewhere).



            Now you can define the induced connection $nabla'$ as follows:



            $$(nabla'_Xpsi)_p=(nabla_tilde{X}tilde psi)_p $$



            where $tilde{X}$ is any extension of $X_p$ and $tilde psi$ is any local extension of $psi$. By the above this is independent of the choice of $tilde{X}$ and $tilde psi$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              For a connection $nabla$ on a vector bundle $E$ over $M$ one can show two things:




              1. If two sections $X,Y$ of $TM$ agree at a point $p$ then $(nabla_Xpsi)_p=(nabla_Ypsi)_p$ for all sections $psi$ of $E$.


              2. If two sections $phi,psi$ agree locally arround $p$ then $(nabla_Xphi)_p=(nabla_Xpsi)_p$ for all sections $X$ of $TM$.



              Furthermore:




              1. Any $X_pin T_pM$ can be extended to a section of $TM$.


              2. For any open neighbourhood $U$ of $p$ and any $psiin E_{|U}$ there exists a section $tilde{psi}$ of $E$ which locally arround $p$ agrees with $psi$ (just take a chart multiply by a suitable bump function and set it $0$ elsewhere).



              Now you can define the induced connection $nabla'$ as follows:



              $$(nabla'_Xpsi)_p=(nabla_tilde{X}tilde psi)_p $$



              where $tilde{X}$ is any extension of $X_p$ and $tilde psi$ is any local extension of $psi$. By the above this is independent of the choice of $tilde{X}$ and $tilde psi$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                For a connection $nabla$ on a vector bundle $E$ over $M$ one can show two things:




                1. If two sections $X,Y$ of $TM$ agree at a point $p$ then $(nabla_Xpsi)_p=(nabla_Ypsi)_p$ for all sections $psi$ of $E$.


                2. If two sections $phi,psi$ agree locally arround $p$ then $(nabla_Xphi)_p=(nabla_Xpsi)_p$ for all sections $X$ of $TM$.



                Furthermore:




                1. Any $X_pin T_pM$ can be extended to a section of $TM$.


                2. For any open neighbourhood $U$ of $p$ and any $psiin E_{|U}$ there exists a section $tilde{psi}$ of $E$ which locally arround $p$ agrees with $psi$ (just take a chart multiply by a suitable bump function and set it $0$ elsewhere).



                Now you can define the induced connection $nabla'$ as follows:



                $$(nabla'_Xpsi)_p=(nabla_tilde{X}tilde psi)_p $$



                where $tilde{X}$ is any extension of $X_p$ and $tilde psi$ is any local extension of $psi$. By the above this is independent of the choice of $tilde{X}$ and $tilde psi$.






                share|cite|improve this answer









                $endgroup$



                For a connection $nabla$ on a vector bundle $E$ over $M$ one can show two things:




                1. If two sections $X,Y$ of $TM$ agree at a point $p$ then $(nabla_Xpsi)_p=(nabla_Ypsi)_p$ for all sections $psi$ of $E$.


                2. If two sections $phi,psi$ agree locally arround $p$ then $(nabla_Xphi)_p=(nabla_Xpsi)_p$ for all sections $X$ of $TM$.



                Furthermore:




                1. Any $X_pin T_pM$ can be extended to a section of $TM$.


                2. For any open neighbourhood $U$ of $p$ and any $psiin E_{|U}$ there exists a section $tilde{psi}$ of $E$ which locally arround $p$ agrees with $psi$ (just take a chart multiply by a suitable bump function and set it $0$ elsewhere).



                Now you can define the induced connection $nabla'$ as follows:



                $$(nabla'_Xpsi)_p=(nabla_tilde{X}tilde psi)_p $$



                where $tilde{X}$ is any extension of $X_p$ and $tilde psi$ is any local extension of $psi$. By the above this is independent of the choice of $tilde{X}$ and $tilde psi$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 15 at 16:58









                triitrii

                80817




                80817






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • 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%2fmath.stackexchange.com%2fquestions%2f253880%2finduced-connection-on-open-sets%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

                    Human spaceflight

                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                    張江高科駅