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Equivalent definition of metric compatibility for a connection: what does $nabla g$ mean?
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So the definition I know for metric compatibility is:
$$Xg(Y,Z)=g(nabla_XY,Z)+g(Y,nabla_XZ),$$
which make sense, as $g(Y,Z)$ is a smooth function from the manifold to reals and we think of $X$ as a derivation. So now I read this apparent equivalent definition that says $nabla g=0$. Can someone explain what this means? How can I do $nabla$ of $g$ I thought $g_p$ is an element of $T_p^*Motimes T_p^*M$ at every point. Furthermore after you explain the meaning of this can you show me that these two definitions are indeed equivalent?
riemannian-geometry tensors connections
edited Apr 21 '16 at 12:18
Michael Albanese
61.2k1591289
asked Apr 19 '16 at 14:12
sifsa
376111
add a comment |Â
up vote
5
down vote
favorite
So the definition I know for metric compatibility is:
$$Xg(Y,Z)=g(nabla_XY,Z)+g(Y,nabla_XZ),$$
which make sense, as $g(Y,Z)$ is a smooth function from the manifold to reals and we think of $X$ as a derivation. So now I read this apparent equivalent definition that says $nabla g=0$. Can someone explain what this means? How can I do $nabla$ of $g$ I thought $g_p$ is an element of $T_p^*Motimes T_p^*M$ at every point. Furthermore after you explain the meaning of this can you show me that these two definitions are indeed equivalent?
riemannian-geometry tensors connections
edited Apr 21 '16 at 12:18
Michael Albanese
61.2k1591289
asked Apr 19 '16 at 14:12
sifsa
376111
Which book are you reading? You should try do Carmo's "Riemannian Geometry".
– Moishe Cohen
Apr 19 '16 at 17:13
Do you know how to take the covariant derivative of a tensor?
– Michael Albanese
Apr 20 '16 at 12:26
I know that by definition a connection is a map taking two vector fields.. I dont understand and have not seen how I can apply it to a tensor
– sifsa
Apr 20 '16 at 16:59
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
So the definition I know for metric compatibility is:
$$Xg(Y,Z)=g(nabla_XY,Z)+g(Y,nabla_XZ),$$
which make sense, as $g(Y,Z)$ is a smooth function from the manifold to reals and we think of $X$ as a derivation. So now I read this apparent equivalent definition that says $nabla g=0$. Can someone explain what this means? How can I do $nabla$ of $g$ I thought $g_p$ is an element of $T_p^*Motimes T_p^*M$ at every point. Furthermore after you explain the meaning of this can you show me that these two definitions are indeed equivalent?
riemannian-geometry tensors connections
edited Apr 21 '16 at 12:18
Michael Albanese
61.2k1591289
asked Apr 19 '16 at 14:12
sifsa
376111
So the definition I know for metric compatibility is:
$$Xg(Y,Z)=g(nabla_XY,Z)+g(Y,nabla_XZ),$$
which make sense, as $g(Y,Z)$ is a smooth function from the manifold to reals and we think of $X$ as a derivation. So now I read this apparent equivalent definition that says $nabla g=0$. Can someone explain what this means? How can I do $nabla$ of $g$ I thought $g_p$ is an element of $T_p^*Motimes T_p^*M$ at every point. Furthermore after you explain the meaning of this can you show me that these two definitions are indeed equivalent?
riemannian-geometry tensors connections
edited Apr 21 '16 at 12:18
Michael Albanese
61.2k1591289
asked Apr 19 '16 at 14:12
sifsa
376111
edited Apr 21 '16 at 12:18
Michael Albanese
61.2k1591289
edited Apr 21 '16 at 12:18
Michael Albanese
61.2k1591289
edited Apr 21 '16 at 12:18
Michael Albanese
61.2k1591289
61.2k1591289
asked Apr 19 '16 at 14:12
sifsa
376111
asked Apr 19 '16 at 14:12
sifsa
376111
asked Apr 19 '16 at 14:12
sifsa
376111
376111
Which book are you reading? You should try do Carmo's "Riemannian Geometry".
– Moishe Cohen
Apr 19 '16 at 17:13
Do you know how to take the covariant derivative of a tensor?
– Michael Albanese
Apr 20 '16 at 12:26
I know that by definition a connection is a map taking two vector fields.. I dont understand and have not seen how I can apply it to a tensor
– sifsa
Apr 20 '16 at 16:59
add a comment |Â
Which book are you reading? You should try do Carmo's "Riemannian Geometry".
– Moishe Cohen
Apr 19 '16 at 17:13
Do you know how to take the covariant derivative of a tensor?
– Michael Albanese
Apr 20 '16 at 12:26
I know that by definition a connection is a map taking two vector fields.. I dont understand and have not seen how I can apply it to a tensor
– sifsa
Apr 20 '16 at 16:59
Which book are you reading? You should try do Carmo's "Riemannian Geometry".
– Moishe Cohen
Apr 19 '16 at 17:13
Which book are you reading? You should try do Carmo's "Riemannian Geometry".
– Moishe Cohen
Apr 19 '16 at 17:13
Do you know how to take the covariant derivative of a tensor?
– Michael Albanese
Apr 20 '16 at 12:26
Do you know how to take the covariant derivative of a tensor?
– Michael Albanese
Apr 20 '16 at 12:26
I know that by definition a connection is a map taking two vector fields.. I dont understand and have not seen how I can apply it to a tensor
– sifsa
Apr 20 '16 at 16:59
I know that by definition a connection is a map taking two vector fields.. I dont understand and have not seen how I can apply it to a tensor
– sifsa
Apr 20 '16 at 16:59
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
12
down vote
accepted
It seems you are missing some necessary background, namely, how to extend a connection on $TM$ to all tensor bundles. I will summarise this construction.
Given a connection
beginalign*
nabla : Gamma(TM) times Gamma(TM) &to Gamma(TM)\
(X, Y) &mapsto nabla_XY
endalign*
on $TM$, there is an associated connection (which I will also denote $nabla$) on $T^*M$ given by
beginalign*
nabla : Gamma(TM) times Gamma(T^*M) &to Gamma(T^*M)\
(X, alpha) &mapsto nabla_Xalpha
endalign*
where $(nabla_Xalpha)(Y) := X(alpha(Y)) - alpha(nabla_XY)$. With this definition, together with the definition $nabla_Xf = Xf$ for a smooth function $f$, we see that the following identity holds:
$$nabla_X(alpha(Y)) = (nabla_Xalpha)(Y) + alpha(nabla_XY).$$
More generally, given a $(p, q)$-tensor $T$, we implicitly define the covariant derivative $nabla_XT$, which is again a $(p, q)$-tensor, by the following equation:
beginalign*
nabla_X(T(Y_1, dots, Y_p, alpha_1, dots, alpha_q)) =& (nabla_XT)(Y_1, dots, Y_p, alpha_1, dots, alpha_q)\
&+ sum_i=1^pT(Y_1, dots, nabla_XY_i, dots, Y_p, alpha_1, dots, alpha_q)\
&+ sum_j=1^qT(Y_1, dots, Y_p, alpha_1, dots, nabla_Xalpha_j, dots, alpha_q).qquad (ast)
endalign*
One could instead consider the covariant derivative of $T$ as a $(p+1, q)$-tensor $nabla T$ given by
$$(nabla T)(X, Y_1, dots, Y_p, alpha_1, dots, alpha_q) := (nabla_XT)(Y_1, dots, Y_p, alpha_1, dots, alpha_q).$$
Now, $g$ is a $(2, 0)$-tensor. So if $Y$ and $Z$ are vector fields, $g(Y, Z)$ is a smooth function and hence
$$nabla_X(g(Y, Z)) = X(g(Y, Z)).$$
On the other hand, by $(ast)$,
$$nabla_X(g(Y, Z)) = (nabla_Xg)(Y, Z) + g(nabla_XY, Z) + g(Y, nabla_XZ).$$
Using these two equations, we see that
$$(nabla g)(X, Y, Z) = (nabla_X g)(Y, Z) = X(g(Y, Z)) - g(nabla_XY, Z) - g(Y, nabla_XZ).$$
So we see that $nabla$ is compatible with the metric $g$ if and only if $(nabla g)(X, Y, Z) = 0$ for all vector fields $X, Y, Z$ (i.e. $nabla g = 0$).
answered Apr 21 '16 at 3:00
Michael Albanese
61.2k1591289
add a comment |Â
up vote
0
down vote
It means the same thing, $nabla g=0$ is equivalent to saying that $g$ is parallel relatively to the connection $nabla$ which is equivalent to saying that $Xg(Y,Z)=g(nabla_XY,Z)+g(Y,nabla_X,Z)$.
It is the answer, the covariant derivative of the $n$ tensor $T$ is defined by $nabla_XT(X_1,..,X_n)=X.T(X_1,..,X_n)-sum_i(X_1,..,nabla_XX_i,..,X_n)$
answered Apr 19 '16 at 14:19
Tsemo Aristide
50.9k11143
2
seriously this is not an answer
– sifsa
Apr 19 '16 at 14:31
how can I apply $nabla $ on $g$
– sifsa
Apr 19 '16 at 14:33
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
It seems you are missing some necessary background, namely, how to extend a connection on $TM$ to all tensor bundles. I will summarise this construction.
Given a connection
beginalign*
nabla : Gamma(TM) times Gamma(TM) &to Gamma(TM)\
(X, Y) &mapsto nabla_XY
endalign*
on $TM$, there is an associated connection (which I will also denote $nabla$) on $T^*M$ given by
beginalign*
nabla : Gamma(TM) times Gamma(T^*M) &to Gamma(T^*M)\
(X, alpha) &mapsto nabla_Xalpha
endalign*
where $(nabla_Xalpha)(Y) := X(alpha(Y)) - alpha(nabla_XY)$. With this definition, together with the definition $nabla_Xf = Xf$ for a smooth function $f$, we see that the following identity holds:
$$nabla_X(alpha(Y)) = (nabla_Xalpha)(Y) + alpha(nabla_XY).$$
More generally, given a $(p, q)$-tensor $T$, we implicitly define the covariant derivative $nabla_XT$, which is again a $(p, q)$-tensor, by the following equation:
beginalign*
nabla_X(T(Y_1, dots, Y_p, alpha_1, dots, alpha_q)) =& (nabla_XT)(Y_1, dots, Y_p, alpha_1, dots, alpha_q)\
&+ sum_i=1^pT(Y_1, dots, nabla_XY_i, dots, Y_p, alpha_1, dots, alpha_q)\
&+ sum_j=1^qT(Y_1, dots, Y_p, alpha_1, dots, nabla_Xalpha_j, dots, alpha_q).qquad (ast)
endalign*
One could instead consider the covariant derivative of $T$ as a $(p+1, q)$-tensor $nabla T$ given by
$$(nabla T)(X, Y_1, dots, Y_p, alpha_1, dots, alpha_q) := (nabla_XT)(Y_1, dots, Y_p, alpha_1, dots, alpha_q).$$
Now, $g$ is a $(2, 0)$-tensor. So if $Y$ and $Z$ are vector fields, $g(Y, Z)$ is a smooth function and hence
$$nabla_X(g(Y, Z)) = X(g(Y, Z)).$$
On the other hand, by $(ast)$,
$$nabla_X(g(Y, Z)) = (nabla_Xg)(Y, Z) + g(nabla_XY, Z) + g(Y, nabla_XZ).$$
Using these two equations, we see that
$$(nabla g)(X, Y, Z) = (nabla_X g)(Y, Z) = X(g(Y, Z)) - g(nabla_XY, Z) - g(Y, nabla_XZ).$$
So we see that $nabla$ is compatible with the metric $g$ if and only if $(nabla g)(X, Y, Z) = 0$ for all vector fields $X, Y, Z$ (i.e. $nabla g = 0$).
answered Apr 21 '16 at 3:00
Michael Albanese
61.2k1591289
add a comment |Â
up vote
12
down vote
accepted
It seems you are missing some necessary background, namely, how to extend a connection on $TM$ to all tensor bundles. I will summarise this construction.
Given a connection
beginalign*
nabla : Gamma(TM) times Gamma(TM) &to Gamma(TM)\
(X, Y) &mapsto nabla_XY
endalign*
on $TM$, there is an associated connection (which I will also denote $nabla$) on $T^*M$ given by
beginalign*
nabla : Gamma(TM) times Gamma(T^*M) &to Gamma(T^*M)\
(X, alpha) &mapsto nabla_Xalpha
endalign*
where $(nabla_Xalpha)(Y) := X(alpha(Y)) - alpha(nabla_XY)$. With this definition, together with the definition $nabla_Xf = Xf$ for a smooth function $f$, we see that the following identity holds:
$$nabla_X(alpha(Y)) = (nabla_Xalpha)(Y) + alpha(nabla_XY).$$
More generally, given a $(p, q)$-tensor $T$, we implicitly define the covariant derivative $nabla_XT$, which is again a $(p, q)$-tensor, by the following equation:
beginalign*
nabla_X(T(Y_1, dots, Y_p, alpha_1, dots, alpha_q)) =& (nabla_XT)(Y_1, dots, Y_p, alpha_1, dots, alpha_q)\
&+ sum_i=1^pT(Y_1, dots, nabla_XY_i, dots, Y_p, alpha_1, dots, alpha_q)\
&+ sum_j=1^qT(Y_1, dots, Y_p, alpha_1, dots, nabla_Xalpha_j, dots, alpha_q).qquad (ast)
endalign*
One could instead consider the covariant derivative of $T$ as a $(p+1, q)$-tensor $nabla T$ given by
$$(nabla T)(X, Y_1, dots, Y_p, alpha_1, dots, alpha_q) := (nabla_XT)(Y_1, dots, Y_p, alpha_1, dots, alpha_q).$$
Now, $g$ is a $(2, 0)$-tensor. So if $Y$ and $Z$ are vector fields, $g(Y, Z)$ is a smooth function and hence
$$nabla_X(g(Y, Z)) = X(g(Y, Z)).$$
On the other hand, by $(ast)$,
$$nabla_X(g(Y, Z)) = (nabla_Xg)(Y, Z) + g(nabla_XY, Z) + g(Y, nabla_XZ).$$
Using these two equations, we see that
$$(nabla g)(X, Y, Z) = (nabla_X g)(Y, Z) = X(g(Y, Z)) - g(nabla_XY, Z) - g(Y, nabla_XZ).$$
So we see that $nabla$ is compatible with the metric $g$ if and only if $(nabla g)(X, Y, Z) = 0$ for all vector fields $X, Y, Z$ (i.e. $nabla g = 0$).
answered Apr 21 '16 at 3:00
Michael Albanese
61.2k1591289
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
It seems you are missing some necessary background, namely, how to extend a connection on $TM$ to all tensor bundles. I will summarise this construction.
Given a connection
beginalign*
nabla : Gamma(TM) times Gamma(TM) &to Gamma(TM)\
(X, Y) &mapsto nabla_XY
endalign*
on $TM$, there is an associated connection (which I will also denote $nabla$) on $T^*M$ given by
beginalign*
nabla : Gamma(TM) times Gamma(T^*M) &to Gamma(T^*M)\
(X, alpha) &mapsto nabla_Xalpha
endalign*
where $(nabla_Xalpha)(Y) := X(alpha(Y)) - alpha(nabla_XY)$. With this definition, together with the definition $nabla_Xf = Xf$ for a smooth function $f$, we see that the following identity holds:
$$nabla_X(alpha(Y)) = (nabla_Xalpha)(Y) + alpha(nabla_XY).$$
More generally, given a $(p, q)$-tensor $T$, we implicitly define the covariant derivative $nabla_XT$, which is again a $(p, q)$-tensor, by the following equation:
beginalign*
nabla_X(T(Y_1, dots, Y_p, alpha_1, dots, alpha_q)) =& (nabla_XT)(Y_1, dots, Y_p, alpha_1, dots, alpha_q)\
&+ sum_i=1^pT(Y_1, dots, nabla_XY_i, dots, Y_p, alpha_1, dots, alpha_q)\
&+ sum_j=1^qT(Y_1, dots, Y_p, alpha_1, dots, nabla_Xalpha_j, dots, alpha_q).qquad (ast)
endalign*
One could instead consider the covariant derivative of $T$ as a $(p+1, q)$-tensor $nabla T$ given by
$$(nabla T)(X, Y_1, dots, Y_p, alpha_1, dots, alpha_q) := (nabla_XT)(Y_1, dots, Y_p, alpha_1, dots, alpha_q).$$
Now, $g$ is a $(2, 0)$-tensor. So if $Y$ and $Z$ are vector fields, $g(Y, Z)$ is a smooth function and hence
$$nabla_X(g(Y, Z)) = X(g(Y, Z)).$$
On the other hand, by $(ast)$,
$$nabla_X(g(Y, Z)) = (nabla_Xg)(Y, Z) + g(nabla_XY, Z) + g(Y, nabla_XZ).$$
Using these two equations, we see that
$$(nabla g)(X, Y, Z) = (nabla_X g)(Y, Z) = X(g(Y, Z)) - g(nabla_XY, Z) - g(Y, nabla_XZ).$$
So we see that $nabla$ is compatible with the metric $g$ if and only if $(nabla g)(X, Y, Z) = 0$ for all vector fields $X, Y, Z$ (i.e. $nabla g = 0$).
answered Apr 21 '16 at 3:00
Michael Albanese
61.2k1591289
It seems you are missing some necessary background, namely, how to extend a connection on $TM$ to all tensor bundles. I will summarise this construction.
Given a connection
beginalign*
nabla : Gamma(TM) times Gamma(TM) &to Gamma(TM)\
(X, Y) &mapsto nabla_XY
endalign*
on $TM$, there is an associated connection (which I will also denote $nabla$) on $T^*M$ given by
beginalign*
nabla : Gamma(TM) times Gamma(T^*M) &to Gamma(T^*M)\
(X, alpha) &mapsto nabla_Xalpha
endalign*
where $(nabla_Xalpha)(Y) := X(alpha(Y)) - alpha(nabla_XY)$. With this definition, together with the definition $nabla_Xf = Xf$ for a smooth function $f$, we see that the following identity holds:
$$nabla_X(alpha(Y)) = (nabla_Xalpha)(Y) + alpha(nabla_XY).$$
More generally, given a $(p, q)$-tensor $T$, we implicitly define the covariant derivative $nabla_XT$, which is again a $(p, q)$-tensor, by the following equation:
beginalign*
nabla_X(T(Y_1, dots, Y_p, alpha_1, dots, alpha_q)) =& (nabla_XT)(Y_1, dots, Y_p, alpha_1, dots, alpha_q)\
&+ sum_i=1^pT(Y_1, dots, nabla_XY_i, dots, Y_p, alpha_1, dots, alpha_q)\
&+ sum_j=1^qT(Y_1, dots, Y_p, alpha_1, dots, nabla_Xalpha_j, dots, alpha_q).qquad (ast)
endalign*
One could instead consider the covariant derivative of $T$ as a $(p+1, q)$-tensor $nabla T$ given by
$$(nabla T)(X, Y_1, dots, Y_p, alpha_1, dots, alpha_q) := (nabla_XT)(Y_1, dots, Y_p, alpha_1, dots, alpha_q).$$
Now, $g$ is a $(2, 0)$-tensor. So if $Y$ and $Z$ are vector fields, $g(Y, Z)$ is a smooth function and hence
$$nabla_X(g(Y, Z)) = X(g(Y, Z)).$$
On the other hand, by $(ast)$,
$$nabla_X(g(Y, Z)) = (nabla_Xg)(Y, Z) + g(nabla_XY, Z) + g(Y, nabla_XZ).$$
Using these two equations, we see that
$$(nabla g)(X, Y, Z) = (nabla_X g)(Y, Z) = X(g(Y, Z)) - g(nabla_XY, Z) - g(Y, nabla_XZ).$$
So we see that $nabla$ is compatible with the metric $g$ if and only if $(nabla g)(X, Y, Z) = 0$ for all vector fields $X, Y, Z$ (i.e. $nabla g = 0$).
answered Apr 21 '16 at 3:00
Michael Albanese
61.2k1591289
edited May 2 '16 at 13:48
answered Apr 21 '16 at 3:00
Michael Albanese
61.2k1591289
answered Apr 21 '16 at 3:00
Michael Albanese
61.2k1591289
answered Apr 21 '16 at 3:00
Michael Albanese
61.2k1591289
61.2k1591289
add a comment |Â
add a comment |Â
up vote
0
down vote
It means the same thing, $nabla g=0$ is equivalent to saying that $g$ is parallel relatively to the connection $nabla$ which is equivalent to saying that $Xg(Y,Z)=g(nabla_XY,Z)+g(Y,nabla_X,Z)$.
It is the answer, the covariant derivative of the $n$ tensor $T$ is defined by $nabla_XT(X_1,..,X_n)=X.T(X_1,..,X_n)-sum_i(X_1,..,nabla_XX_i,..,X_n)$
answered Apr 19 '16 at 14:19
Tsemo Aristide
50.9k11143
2
seriously this is not an answer
– sifsa
Apr 19 '16 at 14:31
how can I apply $nabla $ on $g$
– sifsa
Apr 19 '16 at 14:33
add a comment |Â
up vote
0
down vote
It means the same thing, $nabla g=0$ is equivalent to saying that $g$ is parallel relatively to the connection $nabla$ which is equivalent to saying that $Xg(Y,Z)=g(nabla_XY,Z)+g(Y,nabla_X,Z)$.
It is the answer, the covariant derivative of the $n$ tensor $T$ is defined by $nabla_XT(X_1,..,X_n)=X.T(X_1,..,X_n)-sum_i(X_1,..,nabla_XX_i,..,X_n)$
answered Apr 19 '16 at 14:19
Tsemo Aristide
50.9k11143
2
seriously this is not an answer
– sifsa
Apr 19 '16 at 14:31
how can I apply $nabla $ on $g$
– sifsa
Apr 19 '16 at 14:33
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It means the same thing, $nabla g=0$ is equivalent to saying that $g$ is parallel relatively to the connection $nabla$ which is equivalent to saying that $Xg(Y,Z)=g(nabla_XY,Z)+g(Y,nabla_X,Z)$.
It is the answer, the covariant derivative of the $n$ tensor $T$ is defined by $nabla_XT(X_1,..,X_n)=X.T(X_1,..,X_n)-sum_i(X_1,..,nabla_XX_i,..,X_n)$
answered Apr 19 '16 at 14:19
Tsemo Aristide
50.9k11143
It means the same thing, $nabla g=0$ is equivalent to saying that $g$ is parallel relatively to the connection $nabla$ which is equivalent to saying that $Xg(Y,Z)=g(nabla_XY,Z)+g(Y,nabla_X,Z)$.
It is the answer, the covariant derivative of the $n$ tensor $T$ is defined by $nabla_XT(X_1,..,X_n)=X.T(X_1,..,X_n)-sum_i(X_1,..,nabla_XX_i,..,X_n)$
answered Apr 19 '16 at 14:19
Tsemo Aristide
50.9k11143
edited Apr 19 '16 at 14:50
answered Apr 19 '16 at 14:19
Tsemo Aristide
50.9k11143
answered Apr 19 '16 at 14:19
Tsemo Aristide
50.9k11143
answered Apr 19 '16 at 14:19
Tsemo Aristide
50.9k11143
50.9k11143
2
seriously this is not an answer
– sifsa
Apr 19 '16 at 14:31
how can I apply $nabla $ on $g$
– sifsa
Apr 19 '16 at 14:33
add a comment |Â
2
seriously this is not an answer
– sifsa
Apr 19 '16 at 14:31
how can I apply $nabla $ on $g$
– sifsa
Apr 19 '16 at 14:33
2
2
seriously this is not an answer
– sifsa
Apr 19 '16 at 14:31
seriously this is not an answer
– sifsa
Apr 19 '16 at 14:31
how can I apply $nabla $ on $g$
– sifsa
Apr 19 '16 at 14:33
how can I apply $nabla $ on $g$
– sifsa
Apr 19 '16 at 14:33
add a comment |Â
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Which book are you reading? You should try do Carmo's "Riemannian Geometry".
– Moishe Cohen
Apr 19 '16 at 17:13
Do you know how to take the covariant derivative of a tensor?
– Michael Albanese
Apr 20 '16 at 12:26
I know that by definition a connection is a map taking two vector fields.. I dont understand and have not seen how I can apply it to a tensor
– sifsa
Apr 20 '16 at 16:59