Intrinsic definition of Jacobian matrix on manifolds












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$begingroup$


For a vector field $X:mathbb{R}^{n}rightarrowmathbb{R}^{n}$, the Jacobian matrix at $pinmathbb{R}^{3}$ is defined as
$$mathcal{J}_{p}X:=begin{bmatrix}left.frac{partial X^{i}}{partial x^{j}}right|_{p}end{bmatrix}_{i,j=1, cdots ,n}.$$
This seems to be obviously coordinate-dependent. Also, we usually do not talk about Jacobian matrices of vector fields; rather, we talk about Jacobian matrices for mappings into $mathbb{R}^{n}$, not the tangent bundle.



Contrary to my intuition, it turns out that Jacobian matrix of a vector field can be indeed naturally defined in a coordinate-independent manner.
Let $M$ be a smooth manifold and $XinGamma(TM)$ be a vector field. Then since $X:Mrightarrow TM$ is a smooth function into another manifold $TM$, we can think of the derivative $D_{p}X:T_{p}Mrightarrow T_{X(p)}TM$ at $pin M$.



This derivative is a map into the double tangent bundle $TTM$. There is a map called the canonical flip $J:TTMrightarrow TTM$, which basically swaps "space components" for "tangent components." This canonical flip is an isomorphism between two vector bundles on $TM$: first, the canonical bundle projection $pi_{TTM}:TTMrightarrow TM$ and second, the derivative $pi_{TM*}:TTMrightarrow TM$ of the canonical bundle projection $pi_{TM}:TMrightarrow M$.



In $mathbb{R}^{n}$, the $i$th row of the Jacobian matrix $mathcal{J}_{p}X$ is the derivative of the $i$th component of $X$, so for a tangent vector $Y_{p}$, the matrix multiplication $(mathcal{J}_{p}X)Y_{p}$ is nothing but the "directional derivative" of $X$ along $Y$. Actually, this information is fully contained in "tangent components" of $(D_{p}X)Y_{p}$. The "space components" are really just nothing, because $X$ is a section (thus in terms of the "space components," $X$ is just the identity map on $M$). So, if we swap the "space components" for the "tangent components" using the canonical flip $J$, and then push them back to $M$ using the canonical projection $pi_{TM}$, then what we get is exactly the vector obtained by the matrix multiplication of the "Jacobian matrix" of $X$ with $Y$.



As a result, we get an affine connection $mathcal{J}:Gamma(TM)rightarrowGamma(TMotimes T^{*}M)$ given as
$$mathcal{J}_{p}:X_{p}mapsto(Y_{p}mapstopi_{TM*}(J_{X(p)}D_{p}X(Y_{p}))).$$



(It seems that $mathcal{J}$ is related to another notion of "directional derivative", the Lie derivative: $mathcal{L}_{X}Y=[X,Y]=(mathcal{J}Y)X-(mathcal{J}X)Y$)



It seems to me that the construction is sufficiently natural (though complicated), and I'm curious about (1) what's really going on here (I'm not familiar with double tangent bundle...), and (2) any practical usage of this construction if any.



I'm also curious that (3) whether or not our vector calculus definition of divergence, that is, the trace of Jacobian, coincides (or have any relation) with the natural definition of divergence, e.g., that explained in https://en.wikipedia.org/wiki/Divergence#Generalizations.



By the way, by "what's going on here" I mean some explicit calculations plus some geometric intuitions.



Thanks!










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    For a vector field $X:mathbb{R}^{n}rightarrowmathbb{R}^{n}$, the Jacobian matrix at $pinmathbb{R}^{3}$ is defined as
    $$mathcal{J}_{p}X:=begin{bmatrix}left.frac{partial X^{i}}{partial x^{j}}right|_{p}end{bmatrix}_{i,j=1, cdots ,n}.$$
    This seems to be obviously coordinate-dependent. Also, we usually do not talk about Jacobian matrices of vector fields; rather, we talk about Jacobian matrices for mappings into $mathbb{R}^{n}$, not the tangent bundle.



    Contrary to my intuition, it turns out that Jacobian matrix of a vector field can be indeed naturally defined in a coordinate-independent manner.
    Let $M$ be a smooth manifold and $XinGamma(TM)$ be a vector field. Then since $X:Mrightarrow TM$ is a smooth function into another manifold $TM$, we can think of the derivative $D_{p}X:T_{p}Mrightarrow T_{X(p)}TM$ at $pin M$.



    This derivative is a map into the double tangent bundle $TTM$. There is a map called the canonical flip $J:TTMrightarrow TTM$, which basically swaps "space components" for "tangent components." This canonical flip is an isomorphism between two vector bundles on $TM$: first, the canonical bundle projection $pi_{TTM}:TTMrightarrow TM$ and second, the derivative $pi_{TM*}:TTMrightarrow TM$ of the canonical bundle projection $pi_{TM}:TMrightarrow M$.



    In $mathbb{R}^{n}$, the $i$th row of the Jacobian matrix $mathcal{J}_{p}X$ is the derivative of the $i$th component of $X$, so for a tangent vector $Y_{p}$, the matrix multiplication $(mathcal{J}_{p}X)Y_{p}$ is nothing but the "directional derivative" of $X$ along $Y$. Actually, this information is fully contained in "tangent components" of $(D_{p}X)Y_{p}$. The "space components" are really just nothing, because $X$ is a section (thus in terms of the "space components," $X$ is just the identity map on $M$). So, if we swap the "space components" for the "tangent components" using the canonical flip $J$, and then push them back to $M$ using the canonical projection $pi_{TM}$, then what we get is exactly the vector obtained by the matrix multiplication of the "Jacobian matrix" of $X$ with $Y$.



    As a result, we get an affine connection $mathcal{J}:Gamma(TM)rightarrowGamma(TMotimes T^{*}M)$ given as
    $$mathcal{J}_{p}:X_{p}mapsto(Y_{p}mapstopi_{TM*}(J_{X(p)}D_{p}X(Y_{p}))).$$



    (It seems that $mathcal{J}$ is related to another notion of "directional derivative", the Lie derivative: $mathcal{L}_{X}Y=[X,Y]=(mathcal{J}Y)X-(mathcal{J}X)Y$)



    It seems to me that the construction is sufficiently natural (though complicated), and I'm curious about (1) what's really going on here (I'm not familiar with double tangent bundle...), and (2) any practical usage of this construction if any.



    I'm also curious that (3) whether or not our vector calculus definition of divergence, that is, the trace of Jacobian, coincides (or have any relation) with the natural definition of divergence, e.g., that explained in https://en.wikipedia.org/wiki/Divergence#Generalizations.



    By the way, by "what's going on here" I mean some explicit calculations plus some geometric intuitions.



    Thanks!










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      For a vector field $X:mathbb{R}^{n}rightarrowmathbb{R}^{n}$, the Jacobian matrix at $pinmathbb{R}^{3}$ is defined as
      $$mathcal{J}_{p}X:=begin{bmatrix}left.frac{partial X^{i}}{partial x^{j}}right|_{p}end{bmatrix}_{i,j=1, cdots ,n}.$$
      This seems to be obviously coordinate-dependent. Also, we usually do not talk about Jacobian matrices of vector fields; rather, we talk about Jacobian matrices for mappings into $mathbb{R}^{n}$, not the tangent bundle.



      Contrary to my intuition, it turns out that Jacobian matrix of a vector field can be indeed naturally defined in a coordinate-independent manner.
      Let $M$ be a smooth manifold and $XinGamma(TM)$ be a vector field. Then since $X:Mrightarrow TM$ is a smooth function into another manifold $TM$, we can think of the derivative $D_{p}X:T_{p}Mrightarrow T_{X(p)}TM$ at $pin M$.



      This derivative is a map into the double tangent bundle $TTM$. There is a map called the canonical flip $J:TTMrightarrow TTM$, which basically swaps "space components" for "tangent components." This canonical flip is an isomorphism between two vector bundles on $TM$: first, the canonical bundle projection $pi_{TTM}:TTMrightarrow TM$ and second, the derivative $pi_{TM*}:TTMrightarrow TM$ of the canonical bundle projection $pi_{TM}:TMrightarrow M$.



      In $mathbb{R}^{n}$, the $i$th row of the Jacobian matrix $mathcal{J}_{p}X$ is the derivative of the $i$th component of $X$, so for a tangent vector $Y_{p}$, the matrix multiplication $(mathcal{J}_{p}X)Y_{p}$ is nothing but the "directional derivative" of $X$ along $Y$. Actually, this information is fully contained in "tangent components" of $(D_{p}X)Y_{p}$. The "space components" are really just nothing, because $X$ is a section (thus in terms of the "space components," $X$ is just the identity map on $M$). So, if we swap the "space components" for the "tangent components" using the canonical flip $J$, and then push them back to $M$ using the canonical projection $pi_{TM}$, then what we get is exactly the vector obtained by the matrix multiplication of the "Jacobian matrix" of $X$ with $Y$.



      As a result, we get an affine connection $mathcal{J}:Gamma(TM)rightarrowGamma(TMotimes T^{*}M)$ given as
      $$mathcal{J}_{p}:X_{p}mapsto(Y_{p}mapstopi_{TM*}(J_{X(p)}D_{p}X(Y_{p}))).$$



      (It seems that $mathcal{J}$ is related to another notion of "directional derivative", the Lie derivative: $mathcal{L}_{X}Y=[X,Y]=(mathcal{J}Y)X-(mathcal{J}X)Y$)



      It seems to me that the construction is sufficiently natural (though complicated), and I'm curious about (1) what's really going on here (I'm not familiar with double tangent bundle...), and (2) any practical usage of this construction if any.



      I'm also curious that (3) whether or not our vector calculus definition of divergence, that is, the trace of Jacobian, coincides (or have any relation) with the natural definition of divergence, e.g., that explained in https://en.wikipedia.org/wiki/Divergence#Generalizations.



      By the way, by "what's going on here" I mean some explicit calculations plus some geometric intuitions.



      Thanks!










      share|cite|improve this question











      $endgroup$




      For a vector field $X:mathbb{R}^{n}rightarrowmathbb{R}^{n}$, the Jacobian matrix at $pinmathbb{R}^{3}$ is defined as
      $$mathcal{J}_{p}X:=begin{bmatrix}left.frac{partial X^{i}}{partial x^{j}}right|_{p}end{bmatrix}_{i,j=1, cdots ,n}.$$
      This seems to be obviously coordinate-dependent. Also, we usually do not talk about Jacobian matrices of vector fields; rather, we talk about Jacobian matrices for mappings into $mathbb{R}^{n}$, not the tangent bundle.



      Contrary to my intuition, it turns out that Jacobian matrix of a vector field can be indeed naturally defined in a coordinate-independent manner.
      Let $M$ be a smooth manifold and $XinGamma(TM)$ be a vector field. Then since $X:Mrightarrow TM$ is a smooth function into another manifold $TM$, we can think of the derivative $D_{p}X:T_{p}Mrightarrow T_{X(p)}TM$ at $pin M$.



      This derivative is a map into the double tangent bundle $TTM$. There is a map called the canonical flip $J:TTMrightarrow TTM$, which basically swaps "space components" for "tangent components." This canonical flip is an isomorphism between two vector bundles on $TM$: first, the canonical bundle projection $pi_{TTM}:TTMrightarrow TM$ and second, the derivative $pi_{TM*}:TTMrightarrow TM$ of the canonical bundle projection $pi_{TM}:TMrightarrow M$.



      In $mathbb{R}^{n}$, the $i$th row of the Jacobian matrix $mathcal{J}_{p}X$ is the derivative of the $i$th component of $X$, so for a tangent vector $Y_{p}$, the matrix multiplication $(mathcal{J}_{p}X)Y_{p}$ is nothing but the "directional derivative" of $X$ along $Y$. Actually, this information is fully contained in "tangent components" of $(D_{p}X)Y_{p}$. The "space components" are really just nothing, because $X$ is a section (thus in terms of the "space components," $X$ is just the identity map on $M$). So, if we swap the "space components" for the "tangent components" using the canonical flip $J$, and then push them back to $M$ using the canonical projection $pi_{TM}$, then what we get is exactly the vector obtained by the matrix multiplication of the "Jacobian matrix" of $X$ with $Y$.



      As a result, we get an affine connection $mathcal{J}:Gamma(TM)rightarrowGamma(TMotimes T^{*}M)$ given as
      $$mathcal{J}_{p}:X_{p}mapsto(Y_{p}mapstopi_{TM*}(J_{X(p)}D_{p}X(Y_{p}))).$$



      (It seems that $mathcal{J}$ is related to another notion of "directional derivative", the Lie derivative: $mathcal{L}_{X}Y=[X,Y]=(mathcal{J}Y)X-(mathcal{J}X)Y$)



      It seems to me that the construction is sufficiently natural (though complicated), and I'm curious about (1) what's really going on here (I'm not familiar with double tangent bundle...), and (2) any practical usage of this construction if any.



      I'm also curious that (3) whether or not our vector calculus definition of divergence, that is, the trace of Jacobian, coincides (or have any relation) with the natural definition of divergence, e.g., that explained in https://en.wikipedia.org/wiki/Divergence#Generalizations.



      By the way, by "what's going on here" I mean some explicit calculations plus some geometric intuitions.



      Thanks!







      differential-geometry vector-bundles divergence jacobian






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      edited Dec 4 '18 at 20:14







      Junekey Jeon

















      asked Dec 4 '18 at 20:09









      Junekey JeonJunekey Jeon

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