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Matrix associated of an application between tangent spaces
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Let $M$ be a differential manifold and $X$ a vector field over $M$ s.t. $X(p) = 0$ for some $p in M$. Let be $phi_p : T_p(M) to T_p(M)$ given as $$phi_p(v) := [Y,X](p),$$ being $Y$ another vector field of $M$ s.t. $Y(p) = v$.
For $X = sum_{i=1}^nX_ifrac{partial}{partial x_i}$ calculate the associated matrix of $phi_p$ with respect the basis ${(frac{partial}{partial x_i})}_{1 leq i leq m}$.
Can you help me, please?
And another question: Since $phi_p(v) = [Y,X](p) = Y(X(p)) - X(Y(p)) = Y(0) - X(v) = -X(v)$, can we assert that $phi_p$ does not depend of $Y$?
Thanks
differential-geometry manifolds tangent-spaces
edited yesterday
Ethan Bolker
39k543102
asked yesterday
user540275
306
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up vote
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Let $M$ be a differential manifold and $X$ a vector field over $M$ s.t. $X(p) = 0$ for some $p in M$. Let be $phi_p : T_p(M) to T_p(M)$ given as $$phi_p(v) := [Y,X](p),$$ being $Y$ another vector field of $M$ s.t. $Y(p) = v$.
For $X = sum_{i=1}^nX_ifrac{partial}{partial x_i}$ calculate the associated matrix of $phi_p$ with respect the basis ${(frac{partial}{partial x_i})}_{1 leq i leq m}$.
Can you help me, please?
And another question: Since $phi_p(v) = [Y,X](p) = Y(X(p)) - X(Y(p)) = Y(0) - X(v) = -X(v)$, can we assert that $phi_p$ does not depend of $Y$?
Thanks
differential-geometry manifolds tangent-spaces
edited yesterday
Ethan Bolker
39k543102
asked yesterday
user540275
306
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $M$ be a differential manifold and $X$ a vector field over $M$ s.t. $X(p) = 0$ for some $p in M$. Let be $phi_p : T_p(M) to T_p(M)$ given as $$phi_p(v) := [Y,X](p),$$ being $Y$ another vector field of $M$ s.t. $Y(p) = v$.
For $X = sum_{i=1}^nX_ifrac{partial}{partial x_i}$ calculate the associated matrix of $phi_p$ with respect the basis ${(frac{partial}{partial x_i})}_{1 leq i leq m}$.
Can you help me, please?
And another question: Since $phi_p(v) = [Y,X](p) = Y(X(p)) - X(Y(p)) = Y(0) - X(v) = -X(v)$, can we assert that $phi_p$ does not depend of $Y$?
Thanks
differential-geometry manifolds tangent-spaces
edited yesterday
Ethan Bolker
39k543102
asked yesterday
user540275
306
Let $M$ be a differential manifold and $X$ a vector field over $M$ s.t. $X(p) = 0$ for some $p in M$. Let be $phi_p : T_p(M) to T_p(M)$ given as $$phi_p(v) := [Y,X](p),$$ being $Y$ another vector field of $M$ s.t. $Y(p) = v$.
For $X = sum_{i=1}^nX_ifrac{partial}{partial x_i}$ calculate the associated matrix of $phi_p$ with respect the basis ${(frac{partial}{partial x_i})}_{1 leq i leq m}$.
Can you help me, please?
And another question: Since $phi_p(v) = [Y,X](p) = Y(X(p)) - X(Y(p)) = Y(0) - X(v) = -X(v)$, can we assert that $phi_p$ does not depend of $Y$?
Thanks
differential-geometry manifolds tangent-spaces
differential-geometry manifolds tangent-spaces
edited yesterday
Ethan Bolker
39k543102
asked yesterday
user540275
306
edited yesterday
Ethan Bolker
39k543102
asked yesterday
user540275
306
edited yesterday
Ethan Bolker
39k543102
edited yesterday
Ethan Bolker
39k543102
edited yesterday
Ethan Bolker
39k543102
39k543102
asked yesterday
user540275
306
asked yesterday
user540275
306
asked yesterday
user540275
306
306
add a comment |
add a comment |
1 Answer
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0
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Your last computation is wrong (it actually doesn't make any sense to write $Y(X(p))$) and leads me to believe that the definition of $[Y,X]$ is not clear to you. Here is how it goes. Basically, you need to test how vector fields act on smooth functions. Let $fin C^infty(M)$ and let $X=X^ipartial_i$, $Y=Y^jpartial_j$, where by $partial_i$ I denote $partial/partial_{x_i}$ and I use Einstein summation convention for conciseness. Then
$$
begin{split}
[Y,X](f) &= Y(X(f))-X(Y(f)) \
&= Y^jpartial_j(X^ipartial_if)-X^ipartial_i(Y^jpartial_jf) \
&= Y^j(partial_jX^i)partial_if-X^i(partial_iY^j)partial_jf ,
end{split}
$$
because the terms involving second order derivatives of $f$ cancel by Schwarz's theorem.
At point $p$, we have $X_p^i=0$ and $Y_p^j=v^j$, therefore only the first term remains
$$
[Y,X]_p = v^j(partial_jX^i)_ppartial_i .
$$
In particular, this shows that $[Y,X]_p$ only depends on the value of $Y$ at point $p$, which is $Y_p=v$, and not on the actual vector field itself.
The matrix representation can be recovered from $phi_p(v) = phi_p(v)^ipartial_i$ from which
$$
phi_p(v)^i = (partial_jX^i)_p v^j = M^i_j v^j
$$
with $M^i_j=(partial_jX^i)_p$. The transformation $phi_p$ is linear and depends only on the derivatives of $X$ at $p$.
answered yesterday
Federico
1,99158
So you get $[Y,X](f) = sum_{j=1}^nsum_{i=1}^n v_jBigl(frac{partial X_i}{partial x_j} frac{partial}{partial x_i}Bigr)(p).$ And the matix $phi_p(v) = sum_{i=1}^nphi_p(v)_i frac{partial}{partial x_i}$, with $phi_p(v)_i = Bigl(frac{partial}{partial x_j} X_iBigr)(p)v_j$?
– user540275
yesterday
Yes. That is precisely what I wrote, apart from putting the indices in their most common position. Also, you are missing an $f$ in the r.h.s. of your first formula.
– Federico
yesterday
Ok. And $phi_p (v)_i = phi_p(v_i)$? I suppose, since $phi_p$ is linear (since the partial derivations are linears).
– user540275
yesterday
No. $v^i$ is the $i$-th component of $v$ written in the base $(partial_i)_{i=1}^m$, that is, $v=v^ipartial_i$. It doesn't make sense to write $phi_p(v_i)$. $phi_p$ requires a vector, you cannot feed it just a single coordinate. By $phi_p(v)^i$ i mean the $i$-th component of $phi_p(v)$ written in the base $(partial_i)_{i=1}^m$, that is, $phi_p(v)=phi_p(v)^ipartial_i$.
– Federico
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your last computation is wrong (it actually doesn't make any sense to write $Y(X(p))$) and leads me to believe that the definition of $[Y,X]$ is not clear to you. Here is how it goes. Basically, you need to test how vector fields act on smooth functions. Let $fin C^infty(M)$ and let $X=X^ipartial_i$, $Y=Y^jpartial_j$, where by $partial_i$ I denote $partial/partial_{x_i}$ and I use Einstein summation convention for conciseness. Then
$$
begin{split}
[Y,X](f) &= Y(X(f))-X(Y(f)) \
&= Y^jpartial_j(X^ipartial_if)-X^ipartial_i(Y^jpartial_jf) \
&= Y^j(partial_jX^i)partial_if-X^i(partial_iY^j)partial_jf ,
end{split}
$$
because the terms involving second order derivatives of $f$ cancel by Schwarz's theorem.
At point $p$, we have $X_p^i=0$ and $Y_p^j=v^j$, therefore only the first term remains
$$
[Y,X]_p = v^j(partial_jX^i)_ppartial_i .
$$
In particular, this shows that $[Y,X]_p$ only depends on the value of $Y$ at point $p$, which is $Y_p=v$, and not on the actual vector field itself.
The matrix representation can be recovered from $phi_p(v) = phi_p(v)^ipartial_i$ from which
$$
phi_p(v)^i = (partial_jX^i)_p v^j = M^i_j v^j
$$
with $M^i_j=(partial_jX^i)_p$. The transformation $phi_p$ is linear and depends only on the derivatives of $X$ at $p$.
answered yesterday
Federico
1,99158
So you get $[Y,X](f) = sum_{j=1}^nsum_{i=1}^n v_jBigl(frac{partial X_i}{partial x_j} frac{partial}{partial x_i}Bigr)(p).$ And the matix $phi_p(v) = sum_{i=1}^nphi_p(v)_i frac{partial}{partial x_i}$, with $phi_p(v)_i = Bigl(frac{partial}{partial x_j} X_iBigr)(p)v_j$?
– user540275
yesterday
Yes. That is precisely what I wrote, apart from putting the indices in their most common position. Also, you are missing an $f$ in the r.h.s. of your first formula.
– Federico
yesterday
Ok. And $phi_p (v)_i = phi_p(v_i)$? I suppose, since $phi_p$ is linear (since the partial derivations are linears).
– user540275
yesterday
No. $v^i$ is the $i$-th component of $v$ written in the base $(partial_i)_{i=1}^m$, that is, $v=v^ipartial_i$. It doesn't make sense to write $phi_p(v_i)$. $phi_p$ requires a vector, you cannot feed it just a single coordinate. By $phi_p(v)^i$ i mean the $i$-th component of $phi_p(v)$ written in the base $(partial_i)_{i=1}^m$, that is, $phi_p(v)=phi_p(v)^ipartial_i$.
– Federico
yesterday
add a comment |
up vote
0
down vote
Your last computation is wrong (it actually doesn't make any sense to write $Y(X(p))$) and leads me to believe that the definition of $[Y,X]$ is not clear to you. Here is how it goes. Basically, you need to test how vector fields act on smooth functions. Let $fin C^infty(M)$ and let $X=X^ipartial_i$, $Y=Y^jpartial_j$, where by $partial_i$ I denote $partial/partial_{x_i}$ and I use Einstein summation convention for conciseness. Then
$$
begin{split}
[Y,X](f) &= Y(X(f))-X(Y(f)) \
&= Y^jpartial_j(X^ipartial_if)-X^ipartial_i(Y^jpartial_jf) \
&= Y^j(partial_jX^i)partial_if-X^i(partial_iY^j)partial_jf ,
end{split}
$$
because the terms involving second order derivatives of $f$ cancel by Schwarz's theorem.
At point $p$, we have $X_p^i=0$ and $Y_p^j=v^j$, therefore only the first term remains
$$
[Y,X]_p = v^j(partial_jX^i)_ppartial_i .
$$
In particular, this shows that $[Y,X]_p$ only depends on the value of $Y$ at point $p$, which is $Y_p=v$, and not on the actual vector field itself.
The matrix representation can be recovered from $phi_p(v) = phi_p(v)^ipartial_i$ from which
$$
phi_p(v)^i = (partial_jX^i)_p v^j = M^i_j v^j
$$
with $M^i_j=(partial_jX^i)_p$. The transformation $phi_p$ is linear and depends only on the derivatives of $X$ at $p$.
answered yesterday
Federico
1,99158
So you get $[Y,X](f) = sum_{j=1}^nsum_{i=1}^n v_jBigl(frac{partial X_i}{partial x_j} frac{partial}{partial x_i}Bigr)(p).$ And the matix $phi_p(v) = sum_{i=1}^nphi_p(v)_i frac{partial}{partial x_i}$, with $phi_p(v)_i = Bigl(frac{partial}{partial x_j} X_iBigr)(p)v_j$?
– user540275
yesterday
Yes. That is precisely what I wrote, apart from putting the indices in their most common position. Also, you are missing an $f$ in the r.h.s. of your first formula.
– Federico
yesterday
Ok. And $phi_p (v)_i = phi_p(v_i)$? I suppose, since $phi_p$ is linear (since the partial derivations are linears).
– user540275
yesterday
No. $v^i$ is the $i$-th component of $v$ written in the base $(partial_i)_{i=1}^m$, that is, $v=v^ipartial_i$. It doesn't make sense to write $phi_p(v_i)$. $phi_p$ requires a vector, you cannot feed it just a single coordinate. By $phi_p(v)^i$ i mean the $i$-th component of $phi_p(v)$ written in the base $(partial_i)_{i=1}^m$, that is, $phi_p(v)=phi_p(v)^ipartial_i$.
– Federico
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
Your last computation is wrong (it actually doesn't make any sense to write $Y(X(p))$) and leads me to believe that the definition of $[Y,X]$ is not clear to you. Here is how it goes. Basically, you need to test how vector fields act on smooth functions. Let $fin C^infty(M)$ and let $X=X^ipartial_i$, $Y=Y^jpartial_j$, where by $partial_i$ I denote $partial/partial_{x_i}$ and I use Einstein summation convention for conciseness. Then
$$
begin{split}
[Y,X](f) &= Y(X(f))-X(Y(f)) \
&= Y^jpartial_j(X^ipartial_if)-X^ipartial_i(Y^jpartial_jf) \
&= Y^j(partial_jX^i)partial_if-X^i(partial_iY^j)partial_jf ,
end{split}
$$
because the terms involving second order derivatives of $f$ cancel by Schwarz's theorem.
At point $p$, we have $X_p^i=0$ and $Y_p^j=v^j$, therefore only the first term remains
$$
[Y,X]_p = v^j(partial_jX^i)_ppartial_i .
$$
In particular, this shows that $[Y,X]_p$ only depends on the value of $Y$ at point $p$, which is $Y_p=v$, and not on the actual vector field itself.
The matrix representation can be recovered from $phi_p(v) = phi_p(v)^ipartial_i$ from which
$$
phi_p(v)^i = (partial_jX^i)_p v^j = M^i_j v^j
$$
with $M^i_j=(partial_jX^i)_p$. The transformation $phi_p$ is linear and depends only on the derivatives of $X$ at $p$.
answered yesterday
Federico
1,99158
Your last computation is wrong (it actually doesn't make any sense to write $Y(X(p))$) and leads me to believe that the definition of $[Y,X]$ is not clear to you. Here is how it goes. Basically, you need to test how vector fields act on smooth functions. Let $fin C^infty(M)$ and let $X=X^ipartial_i$, $Y=Y^jpartial_j$, where by $partial_i$ I denote $partial/partial_{x_i}$ and I use Einstein summation convention for conciseness. Then
$$
begin{split}
[Y,X](f) &= Y(X(f))-X(Y(f)) \
&= Y^jpartial_j(X^ipartial_if)-X^ipartial_i(Y^jpartial_jf) \
&= Y^j(partial_jX^i)partial_if-X^i(partial_iY^j)partial_jf ,
end{split}
$$
because the terms involving second order derivatives of $f$ cancel by Schwarz's theorem.
At point $p$, we have $X_p^i=0$ and $Y_p^j=v^j$, therefore only the first term remains
$$
[Y,X]_p = v^j(partial_jX^i)_ppartial_i .
$$
In particular, this shows that $[Y,X]_p$ only depends on the value of $Y$ at point $p$, which is $Y_p=v$, and not on the actual vector field itself.
The matrix representation can be recovered from $phi_p(v) = phi_p(v)^ipartial_i$ from which
$$
phi_p(v)^i = (partial_jX^i)_p v^j = M^i_j v^j
$$
with $M^i_j=(partial_jX^i)_p$. The transformation $phi_p$ is linear and depends only on the derivatives of $X$ at $p$.
answered yesterday
Federico
1,99158
edited yesterday
answered yesterday
Federico
1,99158
answered yesterday
Federico
1,99158
answered yesterday
Federico
1,99158
1,99158
So you get $[Y,X](f) = sum_{j=1}^nsum_{i=1}^n v_jBigl(frac{partial X_i}{partial x_j} frac{partial}{partial x_i}Bigr)(p).$ And the matix $phi_p(v) = sum_{i=1}^nphi_p(v)_i frac{partial}{partial x_i}$, with $phi_p(v)_i = Bigl(frac{partial}{partial x_j} X_iBigr)(p)v_j$?
– user540275
yesterday
Yes. That is precisely what I wrote, apart from putting the indices in their most common position. Also, you are missing an $f$ in the r.h.s. of your first formula.
– Federico
yesterday
Ok. And $phi_p (v)_i = phi_p(v_i)$? I suppose, since $phi_p$ is linear (since the partial derivations are linears).
– user540275
yesterday
No. $v^i$ is the $i$-th component of $v$ written in the base $(partial_i)_{i=1}^m$, that is, $v=v^ipartial_i$. It doesn't make sense to write $phi_p(v_i)$. $phi_p$ requires a vector, you cannot feed it just a single coordinate. By $phi_p(v)^i$ i mean the $i$-th component of $phi_p(v)$ written in the base $(partial_i)_{i=1}^m$, that is, $phi_p(v)=phi_p(v)^ipartial_i$.
– Federico
yesterday
add a comment |
So you get $[Y,X](f) = sum_{j=1}^nsum_{i=1}^n v_jBigl(frac{partial X_i}{partial x_j} frac{partial}{partial x_i}Bigr)(p).$ And the matix $phi_p(v) = sum_{i=1}^nphi_p(v)_i frac{partial}{partial x_i}$, with $phi_p(v)_i = Bigl(frac{partial}{partial x_j} X_iBigr)(p)v_j$?
– user540275
yesterday
Yes. That is precisely what I wrote, apart from putting the indices in their most common position. Also, you are missing an $f$ in the r.h.s. of your first formula.
– Federico
yesterday
Ok. And $phi_p (v)_i = phi_p(v_i)$? I suppose, since $phi_p$ is linear (since the partial derivations are linears).
– user540275
yesterday
No. $v^i$ is the $i$-th component of $v$ written in the base $(partial_i)_{i=1}^m$, that is, $v=v^ipartial_i$. It doesn't make sense to write $phi_p(v_i)$. $phi_p$ requires a vector, you cannot feed it just a single coordinate. By $phi_p(v)^i$ i mean the $i$-th component of $phi_p(v)$ written in the base $(partial_i)_{i=1}^m$, that is, $phi_p(v)=phi_p(v)^ipartial_i$.
– Federico
yesterday
So you get $[Y,X](f) = sum_{j=1}^nsum_{i=1}^n v_jBigl(frac{partial X_i}{partial x_j} frac{partial}{partial x_i}Bigr)(p).$ And the matix $phi_p(v) = sum_{i=1}^nphi_p(v)_i frac{partial}{partial x_i}$, with $phi_p(v)_i = Bigl(frac{partial}{partial x_j} X_iBigr)(p)v_j$?
– user540275
yesterday
So you get $[Y,X](f) = sum_{j=1}^nsum_{i=1}^n v_jBigl(frac{partial X_i}{partial x_j} frac{partial}{partial x_i}Bigr)(p).$ And the matix $phi_p(v) = sum_{i=1}^nphi_p(v)_i frac{partial}{partial x_i}$, with $phi_p(v)_i = Bigl(frac{partial}{partial x_j} X_iBigr)(p)v_j$?
– user540275
yesterday
Yes. That is precisely what I wrote, apart from putting the indices in their most common position. Also, you are missing an $f$ in the r.h.s. of your first formula.
– Federico
yesterday
Yes. That is precisely what I wrote, apart from putting the indices in their most common position. Also, you are missing an $f$ in the r.h.s. of your first formula.
– Federico
yesterday
Ok. And $phi_p (v)_i = phi_p(v_i)$? I suppose, since $phi_p$ is linear (since the partial derivations are linears).
– user540275
yesterday
Ok. And $phi_p (v)_i = phi_p(v_i)$? I suppose, since $phi_p$ is linear (since the partial derivations are linears).
– user540275
yesterday
No. $v^i$ is the $i$-th component of $v$ written in the base $(partial_i)_{i=1}^m$, that is, $v=v^ipartial_i$. It doesn't make sense to write $phi_p(v_i)$. $phi_p$ requires a vector, you cannot feed it just a single coordinate. By $phi_p(v)^i$ i mean the $i$-th component of $phi_p(v)$ written in the base $(partial_i)_{i=1}^m$, that is, $phi_p(v)=phi_p(v)^ipartial_i$.
– Federico
yesterday
No. $v^i$ is the $i$-th component of $v$ written in the base $(partial_i)_{i=1}^m$, that is, $v=v^ipartial_i$. It doesn't make sense to write $phi_p(v_i)$. $phi_p$ requires a vector, you cannot feed it just a single coordinate. By $phi_p(v)^i$ i mean the $i$-th component of $phi_p(v)$ written in the base $(partial_i)_{i=1}^m$, that is, $phi_p(v)=phi_p(v)^ipartial_i$.
– Federico
yesterday
add a comment |
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