Question of Cartan's structure equation (in Principal bundle)

[lichen1983312]

I have question about the proof of Cartan's structure equation in the context of pincipal bundle in Nakahara's book. The attached image is taken from the book.

To show that the curvature two-form $\Omega$ satisfies

$\Omega (X,Y) = {d_P}\omega (X,Y) + [\omega (X),\omega (Y)]$

The author divided the proof into three cases: i, $X,Y \in {H_u}P$, ii. $X \in {H_u}P$ and $Y \in {V_u}P$, iii. $X,Y \in {V_u}P$.

My problem is about the second case, the author said "Since $Y \in {V_u}P$, there is an element $V \in\mathfrak{g}$ such that $Y = {V^\# }$ ($V^\#$ being the fundamental vector field generated by $V$) then $\omega (Y) = V$ is constant and hence $X\omega (Y) = X \cdot V = 0$"

If I look at the context $Y$ is only required to be vertical such that ${\left. Y \right|_u} \in {V_u}P$, so $\omega (Y)$ should depend on $u \in P$ such that ${\omega _\mu }(Y) \ne {\omega _\nu }(Y)$ in general. Am I right? Or the author is just trying to say that $\omega (Y)$ must be constant along $X \in {H_u}P$? If this is the case then how to show it?
Thanks very much

 

[lavinia]

$dω(X_{p},Y_{p}) = X_{p}⋅ω(Y)-Y_{p}⋅ω(X)-ω([X,Y]_{p})$ where $X$ and $Y$ are extensions of $X_{p}$ and $Y_{p}$ to vector fields in a neighborhood of $p$. This formula is independent of the extensions. In particular if $Y_{p}$ is vertical then $Y$ can be chosen to be the vertical vector field determined by the action of the Lie group $G$ on the fiber. By the definition of $ω$, $ω(Y)$ is a constant vector in the tangent space at the identity of $G$.

Notes:

- Since $X_{p}$ is horizontal it can be extended to a horizontal vector field via the action of $G$. With the extensions of $X_{p}$ and $Y_{p}$ to horizontal and vertical vector fields respectively, the first two terms in the formula for $dω$ are zero.

- To compute the curvature 2-form one evaluates $dω$ on the horizontal projections of $X_{p}$ and $Y_{p}$. Since $Y_{p}$ is vertical, the curvature 2 form evaluates to zero in this case.

- In the general case $Ω(X,Y) = dω(hX,hY)$ where $h$ projects the tangent space onto the horizontal subspace.
$dω(hX,hY)=(hX)⋅ω(hY)-(hY)⋅ω(hX)-ω([hX,hY])$. The first two terms are zero.

One gets
$Ω(X,Y) = ω([hX,hY])$ which shows that the curvature detects whether the horizontal distribution is involutive.

 

[lichen1983312]

Thanks very much for answering my question. I am not sure the concept of "extension" here, and I am trying to understand it now. At the same time, is it possible to explain it without talking about "extensions"?

 

[lichen1983312]

$dω(X_{p},Y_{p}) = X_{p}⋅ω(Y)-Y_{p}⋅ω(X)-ω([X,Y]_{p})$ where $X$ and $Y$ are extensions of $X_{p}$ and $Y_{p}$ to vector fields in a neighborhood of $p$. This formula is independent of the extensions. In particular if $Y_{p}$ is vertical then $Y$ can be chosen to be the vertical vector field determined by the action of the Lie group $G$ on the fiber. By the definition of $ω$, $ω(Y)$ is a constant vector in the tangent space at the identity of $G$.

Notes:

- Since $X_{p}$ is horizontal it can be extended to a horizontal vector field via the action of $G$. With the extensions of $X_{p}$ and $Y_{p}$ to horizontal and vertical vector fields respectively, the first two terms in the formula for $dω$ are zero.

- To compute the curvature 2-form one evaluates $dω$ on the horizontal projections of $X_{p}$ and $Y_{p}$. Since $Y_{p}$ is vertical, the curvature 2 form evaluates to zero in this case.

- In the general case $Ω(X,Y) = dω(hX,hY)$ where $h$ projects the tangent space onto the horizontal subspace.
$dω(hX,hY)=(hX)⋅ω(hY)-(hY)⋅ω(hX)-ω([hX,hY])$. The first two terms are zero.

One gets
$Ω(X,Y) = ω([hX,hY])$ which shows that the curvature detects whether the horizontal distribution is involutive.

If possible, can you point out some sources that explain what do you mean by "extension" here ?

 

[lavinia]

Thanks very much for answering my question. I am not sure the concept of "extension" here, and I am trying to understand it now. At the same time, is it possible to explain it without talking about "extensions"?

It just means that the vector field $X$ equals $X_{p}$ at the point $p$. In this sense it extends $X_{p}$ to a vector field. Same idea for $Y$.

 

[lichen1983312]

It just means that the vector field $X$ equals $X_{p}$ at the point $p$. In this sense it extends $X_{p}$ to a vector field.Same idea for $Y$.

Then when you say "independent of extension" do you mean we only require ${\left. Y \right|_u} = {\left. {{V^\# }} \right|_u}$ such that $Y$ and ${V^\# }$ only agree at the point $u$?

but we don't need $Y = {V^\# }$ in a neibeighood of $u$ ?

 

[lavinia]

Then when you say "independent of extension" do you mean we only require ${\left. Y \right|_u} = {\left. {{V^\# }} \right|_u}$ such that $Y$ and ${V^\# }$ only agree at the point $u$?

but we don't need $Y = {V^\# }$ in a neibeighood of $u$ ?

The 2-form $dω$ takes tangent vectors at a point as its arguments, not vector fields. So the expression $dω(X_{p},Y_{p})$ only depends on the two vectors at the point $p$. But if $X$ and $Y$ are vector fields that extend $X_{p}$ and $Y_{p}$ then $dω$ can be written by the expression that you find in your textbook:

$X_{p}⋅ω(Y) - Y_{p}⋅ω(X) - ω([X,Y]_{p})$

One might ask whether different extensions of $X_{p}$ and $Y_{p}$ would give a different answer. It turns out that the answer is always the same. That is: the answer is independent of the extension. This BTW: is a general expression for the exterior derivative of a 1 form not just the connection 1 form.

Suppose for instance that $ω=df$ for some smooth funcion $f$. Then the expression is easily seen to be zero. On the other hand, $dω=d^{2}f =0$ since the interated exterior derivative is always zero.

Try showing that in general the expression gives a differential form.

 

[lichen1983312]

The 2-form $dω$ takes tangent vectors at a point as its arguments, not vector fields. So the expression $dω(X_{p},Y_{p})$ only depends on the two vectors at the point $p$. But if $X$ and $Y$ are vector fields that extend $X_{p}$ and $Y_{p}$ then $dω$ can be written by the expression that you find in your textbook:

$X_{p}⋅ω(Y) - Y_{p}⋅ω(X) - ω([X,Y]_{p})$

One might ask whether different extensions of $X_{p}$ and $Y_{p}$ would give a different answer. It turns out that the answer is always the same. That is: the answer is independent of the extension. This BTW: is a general expression for the exterior derivative of a 1 form not just the connection 1 form.

Suppose for instance that $ω=df$ for some smooth funcion $f$. Then The expression is easily seen to be zero. On the other hand, $dω=d^{2}f =0$ since the interated exterior derivative is always zero.

Try showing that in general the expression gives a differential form.

Thanks for answering it in such a nice way, but if I look at the proof I posted here, there is still something I don't feel so right. Please help me to find out what is wrong.

Before I start, I like to put a explicit derivative of the relationship $d\omega (X,Y) = X[\omega (Y)] - Y[\omega (X)] - \omega ([X,Y])$ here (in real number valued case ), which I think might help me to explain my question better

let $\omega = {\omega _\mu }d{x^\mu } \in {\Omega ^1}(M)$ and we can find

$d\omega (X,Y) = \frac{{\partial {\omega _\mu }}}{{\partial {x^\nu }}}({X^\nu }{Y^\mu } - {X^\mu }{Y^\nu }) = (X{\omega _\mu }){Y^\mu } - (Y{\omega _\mu }){X^\mu }$

on the other hand

$\begin{array}{l} X[\omega (Y)] - Y[\omega (X)] - \omega ([X,Y])\\ = X\omega (Y) - Y\omega (X) - \omega \left[ {(X{Y^\nu } - Y{X^\nu })\frac{\partial }{{\partial {x^\nu }}}} \right]\\ = X({\omega _\nu }{Y^\nu }) - Y({\omega _\nu }{X^\nu }) - {\omega _\nu }(X{Y^\nu } - Y{X^\nu })\\ = (X{\omega _\nu }){Y^\nu } + {\omega _\nu }(X{Y^\nu }) - (Y{\omega _\nu }){X^\nu } - {\omega _\nu }(Y{X^\nu }) - {\omega _\nu }(X{Y^\nu }) + {\omega _\nu }(Y{X^\nu })\\ = (X{\omega _\nu }){Y^\nu } - (Y{\omega _\nu }){X^\nu } \end{array}$

where terms that would make the result depend on extensions, like ${\omega _\nu }(X{Y^\nu })$, are all canceled out. In other words, the result is independent of the extension is a "combined effect" of all three terms, $X[\omega (Y)]$, $Y[\omega (X)]$ and $\omega ([X,Y])$. Or if the extension changed, although all these three terms change in general, the sum is still the same.

Now, go back to my post on the top, let's look at the step ii where $X \in {H_u}P$ and $Y \in {V_u}P$. I expected a similar thing happen to ${d_P}\omega (X,Y) = X[\omega (Y)] - Y[\omega (X)] - \omega ([X,Y])$ like what is described above . However, here is exactly where my confusion comes from. Here, the second terms $Y[\omega (X)]$ is independent of extension naturally as long as we keep $X \in {H_u}P$. The last term is also naturally independent of extension because $[X,Y] \in {H_u}P$ if $X \in {H_u}P$ and $Y \in {V_u}P$ (which is a lemma proved in the book earlier). what left now is

${d_P}\omega (X,Y) = X[\omega (Y)]$

however, by intuition, there is no obvious reason that $X[\omega (Y)]$ , and therefore ${d_P}\omega (X,Y)$, would be independent of extension, right? There must be something wrong here, please help me to find out.
Thanks very much.

 

[lavinia]

Thanks for answering it in such a nice way, but if I look at the proof I posted here, there is still something I don't feel so right. Please help me to find out what is wrong.

Before I start, I like to put a explicit derivative of the relationship $d\omega (X,Y) = X[\omega (Y)] - Y[\omega (X)] - \omega ([X,Y])$ here (in real number valued case ), which I think might help me to explain my question better

let $\omega = {\omega _\mu }d{x^\mu } \in {\Omega ^1}(M)$ and we can find

$d\omega (X,Y) = \frac{{\partial {\omega _\mu }}}{{\partial {x^\nu }}}({X^\nu }{Y^\mu } - {X^\mu }{Y^\nu }) = (X{\omega _\mu }){Y^\mu } - (Y{\omega _\mu }){X^\mu }$

on the other hand

$\begin{array}{l} X[\omega (Y)] - Y[\omega (X)] - \omega ([X,Y])\\ = X\omega (Y) - Y\omega (X) - \omega \left[ {(X{Y^\nu } - Y{X^\nu })\frac{\partial }{{\partial {x^\nu }}}} \right]\\ = X({\omega _\nu }{Y^\nu }) - Y({\omega _\nu }{X^\nu }) - {\omega _\nu }(X{Y^\nu } - Y{X^\nu })\\ = (X{\omega _\nu }){Y^\nu } + {\omega _\nu }(X{Y^\nu }) - (Y{\omega _\nu }){X^\nu } - {\omega _\nu }(Y{X^\nu }) - {\omega _\nu }(X{Y^\nu }) + {\omega _\nu }(Y{X^\nu })\\ = (X{\omega _\nu }){Y^\nu } - (Y{\omega _\nu }){X^\nu } \end{array}$

where terms that would make the result depend on extensions, like ${\omega _\nu }(X{Y^\nu })$, are all canceled out. In other words, the result is independent of the extension is a "combined effect" of all three terms, $X[\omega (Y)]$, $Y[\omega (X)]$ and $\omega ([X,Y])$. Or if the extension changed, although all these three terms change in general, the sum is still the same.

Now, go back to my post on the top, let's look at the step ii where $X \in {H_u}P$ and $Y \in {V_u}P$. I expected a similar thing happen to ${d_P}\omega (X,Y) = X[\omega (Y)] - Y[\omega (X)] - \omega ([X,Y])$ like what is described above . However, here is exactly where my confusion comes from. Here, the second terms $Y[\omega (X)]$ is independent of extension naturally as long as we keep $X \in {H_u}P$. The last term is also naturally independent of extension because $[X,Y] \in {H_u}P$ if $X \in {H_u}P$ and $Y \in {V_u}P$ (which is a lemma proved in the book earlier). what left now is

${d_P}\omega (X,Y) = X[\omega (Y)]$

however, by intuition, there is no obvious reason that $X[\omega (Y)]$ , and therefore ${d_P}\omega (X,Y)$, would be independent of extension, right? There must be something wrong here, please help me to find out.
Thanks very much.

This was answered in post #2.

Also, the individual terms are not independent of the extension. It is $dω$ that is independent,

 

[lavinia]

I found it helpful to work this out for a simple case, that of a 2 dimensional orientable Riemannian manifold. Then the principal bundle can be taken to be the unit circle bundle with structure group $SO(2)$. The action is by angle rotation. The example is simple because the structure group is abelian, the Lie algebra is trivial, and the connection 1 form can be treated as an ordinary differential form. With a Riemannain metric the action of $SO(2)$ is easy to visualize.

 

[fresh_42]

I found it helpful to work this out for a simple case, that of a 2 dimensional orientable Riemannian manifold. Then the principal bundle can be taken to be the unit circle bundle with structure group $SO(2)$. The action is by angle rotation. The example is simple because the structure group is abelian, the Lie algebra is trivial, and the connection 1 form can be treated as an ordinary differential form. With a Riemannain metric the action of $SO(2)$ is easy to visualize.

The easiest example of a non-trivial Lie algebra is $\{X,Y\,\vert \,[X,Y]=2Y\}$ where $X,Y$ can be represented by $X= \begin{bmatrix}1&&0 \\ 0&& -1\end{bmatrix}$ and $Y= \begin{bmatrix}0&&1 \\ 0 && 0\end{bmatrix}$ with a little bit more interesting paths.
Do you know a Riemann manifold where it is the tangent space. I mean beside the matrix group?

 

[lavinia]

The easiest example of a non-trivial Lie algebra is $\{X,Y\,\vert \,[X,Y]=2Y\}$ where $X,Y$ can be represented by $X= \begin{bmatrix}1&&0 \\ 0&& -1\end{bmatrix}$ and $Y= \begin{bmatrix}0&&1 \\ 0 && 0\end{bmatrix}$ with a little bit more interesting paths.
Do you know a Riemann manifold where it is the tangent space. I mean beside the matrix group?

I do not but I am also ignorant. On a general manifold the Lie algebra of vector fields would seem to be infinite dimensional. For a Lie group one has the right or left invariant vector fields and these can span a finite dimensional subspace. But for a general manifold I don't see a natural finite dimensional subspace.

There is a natural algebra associated to each orientable 2 dimensional Riemannian manifold. It is generated by 3 vector fields that are tangent to the unit tangent circle bundle. All three vector fields are invariant under the action of the structure group SO(2) so are analogous to right invariant vector fields on a Lie group. One of the vector fields is just the unit tangent field to the fiber circle (with positive orientation) and the other two are canonical horizontal vector fields. The bracket rules are

$[H_{1},V] = H_{2}$, $[H_{2},V] = - H_{1}$ and $[H_1,H_2] = -KV$ where K is the Gauss curvature, the $H$'s are the two horizontal vector fields, and $V$ is the unit vertical vector field.

If the Gauss curvature is constant then the algebra is a 3 dimensional Lie algebra over the reals.

 

[lichen1983312]

This was answered in post #2.

Also, the individual terms are not independent of the extension. It is $dω$ that is independent,

This was answered in post #2.

Also, the individual terms are not independent of the extension. It is $dω$ that is independent,

Thanks for helping me so much. I was trying to replying you earlier, but I just did not know what to say. Obviously the situation is, you already made your point clearly and I agree with everything you said. But I still feel something wrong and I was trying to express it to you, however you did not understand what I was saying, which is not your problem since it is always hard to understand something that is wrong with a right mind.....

Anyway what I just said might not be clear neither and English is not even my mother language, what really matters now is that:

I finally figured out !

Here is the story, I was desperate and googling every information that might be helpful, then I found this material
http://www.lepp.cornell.edu/~yz98/notes/A brief introduction to characteristic classes from the differentiable viewpoint.pdf
In page 31, you can find a proposition and note

And the original text of Nakahara is
"Lemma 10.2 Let $X \in {H_u}P$ and $Y \in {V_u}P$, then $[X,Y] \in {H_u}P$ "

See this difference? It is at least a very sloppy expression if it is not wrong, and That is exactly where my confusion comes from. In my previous post I said

"...Now, go back to my post on the top, let's look at the step ii where $X \in {H_u}P$ and $Y \in {V_u}P$. I expected a similar thing happen to ${d_P}\omega (X,Y) = X[\omega (Y)] - Y[\omega (X)] - \omega ([X,Y])$ like what is described above . However, here is exactly where my confusion comes from. Here, the second terms $Y[\omega (X)]$ is independent of extension naturally as long as we keep $X \in {H_u}P$. The last term is also naturally independent of extension because $[X,Y] \in {H_u}P$ if $X \in {H_u}P$ and $Y \in {V_u}P$ (which is a lemma proved in the book earlier). what left now is

${d_P}\omega (X,Y) = X[\omega (Y)]$....."

According to Nakahara, I thought the last term $[X,Y]$ is independent on extension since it is 0 anyway. The second term $Y\omega (X)$ is 0 if $X$ is chosen to be a horizontal vector field. What left is $X[\omega (Y)]$=0, independent of extension, which doesn't make sense.

With the updated information, I realized that the third term only vanish when we choose $Y = {V^\# }$, and which makes the first term vanish at the same time because $\omega (Y)$ is a constant. This make everything consistent now and my problem is solved.

 

[fresh_42]

$[H_{1},V] = H_{2}, [H_{2},V] = - H_{1} , [H_1,H_2] = -KV$ where K is the Gauss curvature, the H's are the two horizontal vector fields, and V is the unit vertical vector field.

If the Gauss curvature is constant then the algebra is a 3 dimensional Lie algebra over the reals.

Yes, this gives basically $SL_2$, resp. $sl_2$. I wondered if there is a better to visualize example than the Lie groups. Of course I could "integrate" my two-dimensional example and get a Lie group. The unit circle is just a bit of a too simple example whereas the groups are difficult to imagine / draw. That's why I asked. Maybe the Möbius band or similar would do and you knew.

 

[lavinia]

Thanks for helping me so much. I was trying to replying you earlier, but I just did not know what to say. Obviously the situation is, you already made your point clearly and I agree with everything you said. But I still feel something wrong and I was trying to express it to you, however you did not understand what I was saying, which is not your problem since it is always hard to understand something that is wrong with a right mind.....

Anyway what I just said might not be clear neither and English is not even my mother language, what really matters now is that:

I finally figured out !

Here is the story, I was desperate and googling every information that might be helpful, then I found this material
http://www.lepp.cornell.edu/~yz98/notes/A brief introduction to characteristic classes from the differentiable viewpoint.pdf
In page 31, you can find a proposition and note
View attachment 106819

And the original text of Nakahara is
"Lemma 10.2 Let $X \in {H_u}P$ and $Y \in {V_u}P$, then $[X,Y] \in {H_u}P$ "

See this difference? It is at least a very sloppy expression if it is not wrong, and That is exactly where my confusion comes from. In my previous post I said

"...Now, go back to my post on the top, let's look at the step ii where $X \in {H_u}P$ and $Y \in {V_u}P$. I expected a similar thing happen to ${d_P}\omega (X,Y) = X[\omega (Y)] - Y[\omega (X)] - \omega ([X,Y])$ like what is described above . However, here is exactly where my confusion comes from. Here, the second terms $Y[\omega (X)]$ is independent of extension naturally as long as we keep $X \in {H_u}P$. The last term is also naturally independent of extension because $[X,Y] \in {H_u}P$ if $X \in {H_u}P$ and $Y \in {V_u}P$ (which is a lemma proved in the book earlier). what left now is

${d_P}\omega (X,Y) = X[\omega (Y)]$....."

According to Nakahara, I thought the last term $[X,Y]$ is independent on extension since it is 0 anyway. The second term $Y\omega (X)$ is 0 if $X$ is chosen to be a horizontal vector field. What left is $X[\omega (Y)]$=0, independent of extension, which doesn't make sense.

With the updated information, I realized that the third term only vanish when we choose $Y = {V^\# }$, and which makes the first term vanish at the same time because $\omega (Y)$ is a constant. This make everything consistent now and my problem is solved.

The key fact is that $dω$ takes on the same value for any extensions of the two vectors to vector fields. This is not true of the individual terms in the formula for $dω$.

 

[lavinia]

Yes, this gives basically $SL_2$, resp. $sl_2$. I wondered if there is a better to visualize example than the Lie groups. Of course I could "integrate" my two-dimensional example and get a Lie group. The unit circle is just a bit of a too simple example whereas the groups are difficult to imagine / draw. That's why I asked. Maybe the Möbius band or similar would do and you knew.

For the case of constant positive Gauss curvature one has a sphere and the tangent circle bundle is diffeomorphic to $SO(3)$. I would guess that the action of $SO(2)$ on the fibers is just group multiplication in $SO(3)$. It would be interesting to look at this on the Hopf fibration as well.

For the case of negative constant Gauss curvature the underlying surface has genus greater than one and the unit circle bundle is not a Lie group.

 

[lichen1983312]

$dω(X_{p},Y_{p}) = X_{p}⋅ω(Y)-Y_{p}⋅ω(X)-ω([X,Y]_{p})$ where $X$ and $Y$ are extensions of $X_{p}$ and $Y_{p}$ to vector fields in a neighborhood of $p$. This formula is independent of the extensions. In particular if $Y_{p}$ is vertical then $Y$ can be chosen to be the vertical vector field determined by the action of the Lie group $G$ on the fiber. By the definition of $ω$, $ω(Y)$ is a constant vector in the tangent space at the identity of $G$.

Notes:

- Since $X_{p}$ is horizontal it can be extended to a horizontal vector field via the action of $G$. With the extensions of $X_{p}$ and $Y_{p}$ to horizontal and vertical vector fields respectively, the first two terms in the formula for $dω$ are zero.

- To compute the curvature 2-form one evaluates $dω$ on the horizontal projections of $X_{p}$ and $Y_{p}$. Since $Y_{p}$ is vertical, the curvature 2 form evaluates to zero in this case.

- In the general case $Ω(X,Y) = dω(hX,hY)$ where $h$ projects the tangent space onto the horizontal subspace.
$dω(hX,hY)=(hX)⋅ω(hY)-(hY)⋅ω(hX)-ω([hX,hY])$. The first two terms are zero.

One gets
$Ω(X,Y) = ω([hX,hY])$ which shows that the curvature detects whether the horizontal distribution is involutive.

Sorry, I have to bother you again, if we look at $dω(X_{p},Y_{p}) = X_{p}⋅ω(Y)-Y_{p}⋅ω(X)-ω([X,Y]_{p})$, in the case ${X_p},{Y_p} \in {H_u}P$, the first two terms only vanish when the extension $X$ and $Y$ are are both chosen to be a horizontal fields, right?

then like you said in this case

$\Omega ({X_p},{Y_p}) = - \omega ({[hX,hY]_p}) = - \omega ({[X,Y]_p})$
here ${[X,Y]_p}$ is a function of two fields $X$ and $Y$ and $- \omega ({[X,Y]_p})$ is dependent on the extensions right? but the left hand side $\Omega ({X_p},{Y_p})$ is by definition independent of extensions, is this against intuition?
or as long as we choose $X$ and $Y$ to be horizontal fields, $\omega ({[X,Y]_p})$ always give the same value?

 

[lavinia]

Sorry, I have to bother you again, if we look at $dω(X_{p},Y_{p}) = X_{p}⋅ω(Y)-Y_{p}⋅ω(X)-ω([X,Y]_{p})$, in the case ${X_p},{Y_p} \in {H_u}P$, the first two terms only vanish when the extension $X$ and $Y$ are are both chosen to be a horizontal fields, right?

yes. The bracket product of two horizontal vector fields may not be horizontal. In fact it might be purely vertical.

 

[lichen1983312]

yes

Then how to show that $\omega ({[X,Y]_p})$ is a constant if $X$ and $Y$ are arbitrary horizontal field ?

 

[lavinia]

Then how to show that $\omega ({[X,Y]_p})$ is a constant if $X$ and $Y$ are arbitrary horizontal field ?

I suggest you prove to yourself that the expression for $dω$ in terms of vector fields is a differential form.

 

[lavinia]

Yes, this gives basically $SL_2$, resp. $sl_2$. I wondered if there is a better to visualize example than the Lie groups. Of course I could "integrate" my two-dimensional example and get a Lie group. The unit circle is just a bit of a too simple example whereas the groups are difficult to imagine / draw. That's why I asked. Maybe the Möbius band or similar would do and you knew.

I think for constant Gauss curvature 1 one gets the Lie algebra of $SO(3)$

This must be true since for the unit sphere since the principal circle bundle is naturally isomorphic $SO(3)$.

According to Wikipedia, the Lie algebra of $SO(3)$ resp. $SU(2)$ can be calculated from Pauli matrices as $[t_{i},t_{j}] = ε_{ijk}t_{k}$

https://en.wikipedia.org/wiki/Rotation_group_SO(3)#A_note_on_Lie_algebra

 

[fresh_42]

I think for constant Gauss curvature 1 one gets the Lie algebra of $SO(3)$

This must be true since for the unit sphere since the principal circle bundle is naturally isomorphic $SO(3)$.

According to Wikipedia, the Lie algebra of $SO(3)$ resp. $SU(2)$ can be calculated from Pauli matrices as $[t_{i},t_{j}] = ε_{ijk}t_{k}$

https://en.wikipedia.org/wiki/Rotation_group_SO(3)#A_note_on_Lie_algebra

Yes, however, from an algebraic standpoint, there is only one (complex) three-dimensional simple Lie algebra - up to isomorphisms.
That's why I wrote "basically" and $sl_2$ for short. The classification of representations of $sl_2$ are often used as an example of the general case in textbooks. Therefore the representations of $SU(2)\, , \,su(2)$ come in for free. (I'm just mentioning this, for there are often questions about them on PF).

 

[lavinia]

Yes, however, from an algebraic standpoint, there is only one (complex) three-dimensional simple Lie algebra - up to isomorphisms.
That's why I wrote "basically" and $sl_2$ for short. The classification of representations of $sl_2$ are often used as an example of the general case in textbooks. Therefore the representations of $SU(2)\, , \,su(2)$ come in for free. (I'm just mentioning this, for there are often questions about them on PF).

Interesting. Is that easy to prove?

 

[fresh_42]

Interesting. Is that easy to prove?

That there is only one simple (complex) three-dimensional Lie algebra or its representations?
The latter depends on where you start with. The Pauli matrices in their original form
$$t_1=\begin{bmatrix}0&&1 \\ 1&& 0\end{bmatrix}\, , \, t_2=\begin{bmatrix}0&&-i \\ i&& 0\end{bmatrix}\, , \, t_3=\begin{bmatrix}1&&0 \\ 0&& -1\end{bmatrix}$$
already are a basis of $\mathfrak{sl}(2,\mathbb{C})=\{A \in \mathbb{M}(2,\mathbb{C})\, | \, tr(A) = 0\}$.

Shorter would be the statement that its Cartan subalgebra is one dimensional which leaves only one positive and one negative root left.

Or even shorter: The simpliest Dynkin diagram is $\circ$

(I'm not quite sure about the real situation. The multiplications might lead to anti-isomorphic copies.)

 

[Alex_Ky]

The key fact is that $dω$ takes on the same value for any extensions of the two vectors to vector fields. This is not true of the individual terms in the formula for $dω$.

I wonder if this equation is correct: $dω(X, Y) = X[\omega(Y)] - Y[\omega(X)] - \omega([X, Y])$. I know it is correct for regular one-forms and for the Maurer-Cartan form that's entirely in the Lie group. The proofs in all other books that I can find don't seem to use this equation.

Also, Nakahara's other lemma that the bracket of horizontal vector fields $[X^H, Y^H]$ is vertical is also incorrect I think.

 

[lavinia]

Also, Nakahara's other lemma that the bracket of horizontal vector fields $[X^H, Y^H]$ is vertical is also incorrect I think.

For a general connection on a principal bundle the bracket of horizontal vector fields is not necessarily vertical. However, if the connection is not flat(if the curvature 2 form is not zero) there is always some pair of horizontal vector fields whose bracket product has a non-zero vertical component.

I know of a case where the bracket product of horizontal vector fields is always vertical. On 2 dimensional orientable surfaces with a Riemannian metric, the bracket product of horizontal vector fields is vertical for a unique connection on the tangent unit circle bundle (which is a principal SO(2) bundle). This is called the Riemannian connection in Singer and Thorpe's Lecture Notes on Elementary Topology and Geometry.

I would guess that this generalizes to a unique connection on the bundle of tangent oriented orthonormal frames on an n dimensional oriented manifold with a Riemannian metric. I would also guess that these connections correspond to the Levi-Civita connection on a Riemannian manifold and the connections where bracket products of horizontal vector fields have horizontal components correspond to connections with non-zero torsion.

 

[Alex_Ky]

I wonder if this equation is correct: $dω(X, Y) = X[\omega(Y)] - Y[\omega(X)] - \omega([X, Y])$. I know it is correct for regular one-forms and for the Maurer-Cartan form that's entirely in the Lie group. The proofs in all other books that I can find don't seem to use this equation.

Also, Nakahara's other lemma that the bracket of horizontal vector fields $[X^H, Y^H]$ is vertical is also incorrect I think.

Just correcting myself here.
$dω(X, Y) = X[\omega(Y)] - Y[\omega(X)] - \omega([X, Y])$ is correct in general including vector-valued forms, which can be proved by direct computation.
On $[X^H, Y^H]$ is vertical: in general not true but that's not what Nakahara meant. The book started with commuting vector fields in $TM$, that's why the commuter of their lift is vertical.

 

[Alex_Ky]

For a general connection on a principal bundle the bracket of horizontal vector fields is not necessarily vertical. However, if the connection is not flat(if the curvature 2 form is not zero) there is always some pair of horizontal vector fields whose bracket product has a non-zero vertical component.

I know of a case where the bracket product of horizontal vector fields is always vertical. On 2 dimensional orientable surfaces with a Riemannian metric, the bracket product of horizontal vector fields is vertical for a unique connection on the tangent unit circle bundle (which is a principal SO(2) bundle). This is called the Riemannian connection in Singer and Thorpe's Lecture Notes on Elementary Topology and Geometry.

I would guess that this generalizes to a unique connection on the bundle of tangent oriented orthonormal frames on an n dimensional oriented manifold with a Riemannian metric. I would also guess that these connections correspond to the Levi-Civita connection on a Riemannian manifold and the connections where bracket products of horizontal vector fields have horizontal components correspond to connections with non-zero torsion.

Thank you very much!