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I'm following Cartan's Differential forms. I'm trying to do exercise 8 on page 161. The chapter is about moving frames and differential forms in surface theory.
Consider the frame of Ex. 2 (principal frame), show that if $dk_1 = dk_2 = 0$ at the point M, then at M $k_1 = k_2$ or $omega_12 = 0$. Deduce that on a surface S which has constant strictly positive gaussian curvature K, the principal curvature cannot have a relative maximum or minimum at a point which is not umbilical.
For the first part all ok. In fact we have
$omega_13 = k_1omega_1 \ omega_23 = k_2omega_2 \ domega_1 = -omega_2wedgeomega_12 \ domega_2 = omega_1wedgeomega_12 \ domega_12 = -k_1k_2omega_1wedgeomega_2 \ domega_13 = k_2domega_1 \ domega_23 = k_1domega_2$
Differentiating the first two and substituting the last two
$ domega_13 = dk_1wedgeomega_1 + k_1domega_1 = k_2domega_1 \ domega_23 = dk_2wedgeomega_2 + k_2domega_2 = k_1domega_2$
we obtain
$dk_1wedgeomega_1 = (k_2 - k_1)domega_1 \ dk_2wedgeomega_2 = (k_1 - k_2)domega_2 $
So if $dk_1 = dk_2 = 0$ we have or $k_1 = k_2$ or $domega_1 = domega_2 = 0$ and so or $k_1 = k_2$ or $omega_12 = 0$.
Now suppose that M is not umbilical, so $k_1 neq k_2$ and $omega_12 = 0$. The frame becomes at M
$omega_12 = 0 \ omega_13 = k_1omega_1 \ omega_23 = k_2omega_2 \ domega_1 = 0 \ domega_2 = 0 \ domega_12 = -k_1k_2omega_1wedgeomega_2 \ domega_13 = 0 \ domega_23 = 0$
Now I don't know how to continue to get a contradiction. I know I have to show that $k_1k_2 leq 0$, against the hypothesis. I also know the solution working in local coordinates, but I don't know how can I translate this in the language of differential forms. The proof here is from Shifrin's book
I don't know how to get second derivatives with differential forms (because $d^2omega = 0$), but I suppose (and also I prefer) I have to work avoiding local coordinates.
Thanks in advance
differential-geometry differential-forms surfaces
asked Aug 6 at 13:02
Marco All-in Nervo
1128
add a comment |Â
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0
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I'm following Cartan's Differential forms. I'm trying to do exercise 8 on page 161. The chapter is about moving frames and differential forms in surface theory.
Consider the frame of Ex. 2 (principal frame), show that if $dk_1 = dk_2 = 0$ at the point M, then at M $k_1 = k_2$ or $omega_12 = 0$. Deduce that on a surface S which has constant strictly positive gaussian curvature K, the principal curvature cannot have a relative maximum or minimum at a point which is not umbilical.
For the first part all ok. In fact we have
$omega_13 = k_1omega_1 \ omega_23 = k_2omega_2 \ domega_1 = -omega_2wedgeomega_12 \ domega_2 = omega_1wedgeomega_12 \ domega_12 = -k_1k_2omega_1wedgeomega_2 \ domega_13 = k_2domega_1 \ domega_23 = k_1domega_2$
Differentiating the first two and substituting the last two
$ domega_13 = dk_1wedgeomega_1 + k_1domega_1 = k_2domega_1 \ domega_23 = dk_2wedgeomega_2 + k_2domega_2 = k_1domega_2$
we obtain
$dk_1wedgeomega_1 = (k_2 - k_1)domega_1 \ dk_2wedgeomega_2 = (k_1 - k_2)domega_2 $
So if $dk_1 = dk_2 = 0$ we have or $k_1 = k_2$ or $domega_1 = domega_2 = 0$ and so or $k_1 = k_2$ or $omega_12 = 0$.
Now suppose that M is not umbilical, so $k_1 neq k_2$ and $omega_12 = 0$. The frame becomes at M
$omega_12 = 0 \ omega_13 = k_1omega_1 \ omega_23 = k_2omega_2 \ domega_1 = 0 \ domega_2 = 0 \ domega_12 = -k_1k_2omega_1wedgeomega_2 \ domega_13 = 0 \ domega_23 = 0$
Now I don't know how to continue to get a contradiction. I know I have to show that $k_1k_2 leq 0$, against the hypothesis. I also know the solution working in local coordinates, but I don't know how can I translate this in the language of differential forms. The proof here is from Shifrin's book
I don't know how to get second derivatives with differential forms (because $d^2omega = 0$), but I suppose (and also I prefer) I have to work avoiding local coordinates.
Thanks in advance
differential-geometry differential-forms surfaces
asked Aug 6 at 13:02
Marco All-in Nervo
1128
So is Cartan being explicit and saying (as in Hilbert's Lemma, which you've quoted from my text) that the smaller principal curvature has a local minimum and the larger principal curvature has a local maximum? Your statement of the question does not include these details.
– Ted Shifrin
Aug 6 at 21:20
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
I'm following Cartan's Differential forms. I'm trying to do exercise 8 on page 161. The chapter is about moving frames and differential forms in surface theory.
Consider the frame of Ex. 2 (principal frame), show that if $dk_1 = dk_2 = 0$ at the point M, then at M $k_1 = k_2$ or $omega_12 = 0$. Deduce that on a surface S which has constant strictly positive gaussian curvature K, the principal curvature cannot have a relative maximum or minimum at a point which is not umbilical.
For the first part all ok. In fact we have
$omega_13 = k_1omega_1 \ omega_23 = k_2omega_2 \ domega_1 = -omega_2wedgeomega_12 \ domega_2 = omega_1wedgeomega_12 \ domega_12 = -k_1k_2omega_1wedgeomega_2 \ domega_13 = k_2domega_1 \ domega_23 = k_1domega_2$
Differentiating the first two and substituting the last two
$ domega_13 = dk_1wedgeomega_1 + k_1domega_1 = k_2domega_1 \ domega_23 = dk_2wedgeomega_2 + k_2domega_2 = k_1domega_2$
we obtain
$dk_1wedgeomega_1 = (k_2 - k_1)domega_1 \ dk_2wedgeomega_2 = (k_1 - k_2)domega_2 $
So if $dk_1 = dk_2 = 0$ we have or $k_1 = k_2$ or $domega_1 = domega_2 = 0$ and so or $k_1 = k_2$ or $omega_12 = 0$.
Now suppose that M is not umbilical, so $k_1 neq k_2$ and $omega_12 = 0$. The frame becomes at M
$omega_12 = 0 \ omega_13 = k_1omega_1 \ omega_23 = k_2omega_2 \ domega_1 = 0 \ domega_2 = 0 \ domega_12 = -k_1k_2omega_1wedgeomega_2 \ domega_13 = 0 \ domega_23 = 0$
Now I don't know how to continue to get a contradiction. I know I have to show that $k_1k_2 leq 0$, against the hypothesis. I also know the solution working in local coordinates, but I don't know how can I translate this in the language of differential forms. The proof here is from Shifrin's book
I don't know how to get second derivatives with differential forms (because $d^2omega = 0$), but I suppose (and also I prefer) I have to work avoiding local coordinates.
Thanks in advance
differential-geometry differential-forms surfaces
asked Aug 6 at 13:02
Marco All-in Nervo
1128
I'm following Cartan's Differential forms. I'm trying to do exercise 8 on page 161. The chapter is about moving frames and differential forms in surface theory.
Consider the frame of Ex. 2 (principal frame), show that if $dk_1 = dk_2 = 0$ at the point M, then at M $k_1 = k_2$ or $omega_12 = 0$. Deduce that on a surface S which has constant strictly positive gaussian curvature K, the principal curvature cannot have a relative maximum or minimum at a point which is not umbilical.
For the first part all ok. In fact we have
$omega_13 = k_1omega_1 \ omega_23 = k_2omega_2 \ domega_1 = -omega_2wedgeomega_12 \ domega_2 = omega_1wedgeomega_12 \ domega_12 = -k_1k_2omega_1wedgeomega_2 \ domega_13 = k_2domega_1 \ domega_23 = k_1domega_2$
Differentiating the first two and substituting the last two
$ domega_13 = dk_1wedgeomega_1 + k_1domega_1 = k_2domega_1 \ domega_23 = dk_2wedgeomega_2 + k_2domega_2 = k_1domega_2$
we obtain
$dk_1wedgeomega_1 = (k_2 - k_1)domega_1 \ dk_2wedgeomega_2 = (k_1 - k_2)domega_2 $
So if $dk_1 = dk_2 = 0$ we have or $k_1 = k_2$ or $domega_1 = domega_2 = 0$ and so or $k_1 = k_2$ or $omega_12 = 0$.
Now suppose that M is not umbilical, so $k_1 neq k_2$ and $omega_12 = 0$. The frame becomes at M
$omega_12 = 0 \ omega_13 = k_1omega_1 \ omega_23 = k_2omega_2 \ domega_1 = 0 \ domega_2 = 0 \ domega_12 = -k_1k_2omega_1wedgeomega_2 \ domega_13 = 0 \ domega_23 = 0$
Now I don't know how to continue to get a contradiction. I know I have to show that $k_1k_2 leq 0$, against the hypothesis. I also know the solution working in local coordinates, but I don't know how can I translate this in the language of differential forms. The proof here is from Shifrin's book
I don't know how to get second derivatives with differential forms (because $d^2omega = 0$), but I suppose (and also I prefer) I have to work avoiding local coordinates.
Thanks in advance
differential-geometry differential-forms surfaces
asked Aug 6 at 13:02
Marco All-in Nervo
1128
edited Aug 6 at 13:11
asked Aug 6 at 13:02
Marco All-in Nervo
1128
asked Aug 6 at 13:02
Marco All-in Nervo
1128
asked Aug 6 at 13:02
Marco All-in Nervo
1128
1128
So is Cartan being explicit and saying (as in Hilbert's Lemma, which you've quoted from my text) that the smaller principal curvature has a local minimum and the larger principal curvature has a local maximum? Your statement of the question does not include these details.
– Ted Shifrin
Aug 6 at 21:20
add a comment |Â
So is Cartan being explicit and saying (as in Hilbert's Lemma, which you've quoted from my text) that the smaller principal curvature has a local minimum and the larger principal curvature has a local maximum? Your statement of the question does not include these details.
– Ted Shifrin
Aug 6 at 21:20
So is Cartan being explicit and saying (as in Hilbert's Lemma, which you've quoted from my text) that the smaller principal curvature has a local minimum and the larger principal curvature has a local maximum? Your statement of the question does not include these details.
– Ted Shifrin
Aug 6 at 21:20
So is Cartan being explicit and saying (as in Hilbert's Lemma, which you've quoted from my text) that the smaller principal curvature has a local minimum and the larger principal curvature has a local maximum? Your statement of the question does not include these details.
– Ted Shifrin
Aug 6 at 21:20
add a comment |Â
1 Answer
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1
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Even with the moving frames computation, you're going to have to do something analogous to the local computation with second-order partial derivatives. How else can we check that a critical point is a local maximum/minimum?
Here's how you should start: Write $dk_i = sumlimits_j k_ijomega_j$ (so we know that $k_ij = 0$ at $M$ for $i,j=1,2$). Then write $dk_ij = sumlimits_ell k_ijellomega_ell$. If $k_1>k_2$ locally, then we know that $k_1jj le 0$ and $k_2jjge 0$ at $M$ for $j=1,2$.
I would rather write your third displayed equations as
beginalign*
dk_1wedgeomega_1 &= (k_2-k_1)omega_12wedgeomega_2 \
dk_2wedgeomega_2 &= (k_1-k_2)omega_1wedgeomega_12.
endalign*
Solve these to obtain $(k_1-k_2)omega_12 = Aomega_1+Bomega_2$. Now can you proceed?
answered Aug 6 at 21:36
Ted Shifrin
59.6k44387
I get $A = k_12$ and $B = k_21$. Then differentiating your last equation I obtain at M $ -(k_1 - k_2)k_1k_2 = -k_122+k_211$, which is a contradiction since the LHS is $lt 0$ and the RHS is $geq 0$. Thank you very very much! I know I had to take the second derivative, but I didn't know how to do it! The "trick" you used - write the differential $dk_i$ with respect of the dual basis and then differentiate the coefficients $k_ij$ instead of the whole 1-form - is great! Thank you again!
– Marco All-in Nervo
Aug 7 at 14:34
Is this the standard way to get second derivatives using differential forms?
– Marco All-in Nervo
Aug 7 at 14:40
Pretty much, yes. It corresponds locally, if you like, to taking the (co)tangent bundle of the (co)tangent bundle.
– Ted Shifrin
Aug 7 at 16:06
Could you expand this a little bit?
– Marco All-in Nervo
Aug 7 at 18:22
You might start with this, but don't get too carried away. If you haven't thought about vector bundles much, I suggest waiting. Intuitively, I just meant that if you think of $1$-form $omega$ as a section of $T^*X$, then the cotangent space to the fiber of $T^*X$ at $omega(p)=sum a_i(p)dx^i(p)$ will consist (locally) of derivatives of $a_i$ at $p$. You can think about local coordinates for $T^*X$, $n=dim X$ coming from $X$ and $n$ coming from the coefficients of a $1$-form with respect to a basis (in our set up, that was the $omega_i$).
– Ted Shifrin
Aug 7 at 18:30
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Even with the moving frames computation, you're going to have to do something analogous to the local computation with second-order partial derivatives. How else can we check that a critical point is a local maximum/minimum?
Here's how you should start: Write $dk_i = sumlimits_j k_ijomega_j$ (so we know that $k_ij = 0$ at $M$ for $i,j=1,2$). Then write $dk_ij = sumlimits_ell k_ijellomega_ell$. If $k_1>k_2$ locally, then we know that $k_1jj le 0$ and $k_2jjge 0$ at $M$ for $j=1,2$.
I would rather write your third displayed equations as
beginalign*
dk_1wedgeomega_1 &= (k_2-k_1)omega_12wedgeomega_2 \
dk_2wedgeomega_2 &= (k_1-k_2)omega_1wedgeomega_12.
endalign*
Solve these to obtain $(k_1-k_2)omega_12 = Aomega_1+Bomega_2$. Now can you proceed?
answered Aug 6 at 21:36
Ted Shifrin
59.6k44387
I get $A = k_12$ and $B = k_21$. Then differentiating your last equation I obtain at M $ -(k_1 - k_2)k_1k_2 = -k_122+k_211$, which is a contradiction since the LHS is $lt 0$ and the RHS is $geq 0$. Thank you very very much! I know I had to take the second derivative, but I didn't know how to do it! The "trick" you used - write the differential $dk_i$ with respect of the dual basis and then differentiate the coefficients $k_ij$ instead of the whole 1-form - is great! Thank you again!
– Marco All-in Nervo
Aug 7 at 14:34
Is this the standard way to get second derivatives using differential forms?
– Marco All-in Nervo
Aug 7 at 14:40
Pretty much, yes. It corresponds locally, if you like, to taking the (co)tangent bundle of the (co)tangent bundle.
– Ted Shifrin
Aug 7 at 16:06
Could you expand this a little bit?
– Marco All-in Nervo
Aug 7 at 18:22
You might start with this, but don't get too carried away. If you haven't thought about vector bundles much, I suggest waiting. Intuitively, I just meant that if you think of $1$-form $omega$ as a section of $T^*X$, then the cotangent space to the fiber of $T^*X$ at $omega(p)=sum a_i(p)dx^i(p)$ will consist (locally) of derivatives of $a_i$ at $p$. You can think about local coordinates for $T^*X$, $n=dim X$ coming from $X$ and $n$ coming from the coefficients of a $1$-form with respect to a basis (in our set up, that was the $omega_i$).
– Ted Shifrin
Aug 7 at 18:30
 |Â
show 4 more comments
up vote
1
down vote
accepted
Even with the moving frames computation, you're going to have to do something analogous to the local computation with second-order partial derivatives. How else can we check that a critical point is a local maximum/minimum?
Here's how you should start: Write $dk_i = sumlimits_j k_ijomega_j$ (so we know that $k_ij = 0$ at $M$ for $i,j=1,2$). Then write $dk_ij = sumlimits_ell k_ijellomega_ell$. If $k_1>k_2$ locally, then we know that $k_1jj le 0$ and $k_2jjge 0$ at $M$ for $j=1,2$.
I would rather write your third displayed equations as
beginalign*
dk_1wedgeomega_1 &= (k_2-k_1)omega_12wedgeomega_2 \
dk_2wedgeomega_2 &= (k_1-k_2)omega_1wedgeomega_12.
endalign*
Solve these to obtain $(k_1-k_2)omega_12 = Aomega_1+Bomega_2$. Now can you proceed?
answered Aug 6 at 21:36
Ted Shifrin
59.6k44387
I get $A = k_12$ and $B = k_21$. Then differentiating your last equation I obtain at M $ -(k_1 - k_2)k_1k_2 = -k_122+k_211$, which is a contradiction since the LHS is $lt 0$ and the RHS is $geq 0$. Thank you very very much! I know I had to take the second derivative, but I didn't know how to do it! The "trick" you used - write the differential $dk_i$ with respect of the dual basis and then differentiate the coefficients $k_ij$ instead of the whole 1-form - is great! Thank you again!
– Marco All-in Nervo
Aug 7 at 14:34
Is this the standard way to get second derivatives using differential forms?
– Marco All-in Nervo
Aug 7 at 14:40
Pretty much, yes. It corresponds locally, if you like, to taking the (co)tangent bundle of the (co)tangent bundle.
– Ted Shifrin
Aug 7 at 16:06
Could you expand this a little bit?
– Marco All-in Nervo
Aug 7 at 18:22
You might start with this, but don't get too carried away. If you haven't thought about vector bundles much, I suggest waiting. Intuitively, I just meant that if you think of $1$-form $omega$ as a section of $T^*X$, then the cotangent space to the fiber of $T^*X$ at $omega(p)=sum a_i(p)dx^i(p)$ will consist (locally) of derivatives of $a_i$ at $p$. You can think about local coordinates for $T^*X$, $n=dim X$ coming from $X$ and $n$ coming from the coefficients of a $1$-form with respect to a basis (in our set up, that was the $omega_i$).
– Ted Shifrin
Aug 7 at 18:30
 |Â
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Even with the moving frames computation, you're going to have to do something analogous to the local computation with second-order partial derivatives. How else can we check that a critical point is a local maximum/minimum?
Here's how you should start: Write $dk_i = sumlimits_j k_ijomega_j$ (so we know that $k_ij = 0$ at $M$ for $i,j=1,2$). Then write $dk_ij = sumlimits_ell k_ijellomega_ell$. If $k_1>k_2$ locally, then we know that $k_1jj le 0$ and $k_2jjge 0$ at $M$ for $j=1,2$.
I would rather write your third displayed equations as
beginalign*
dk_1wedgeomega_1 &= (k_2-k_1)omega_12wedgeomega_2 \
dk_2wedgeomega_2 &= (k_1-k_2)omega_1wedgeomega_12.
endalign*
Solve these to obtain $(k_1-k_2)omega_12 = Aomega_1+Bomega_2$. Now can you proceed?
answered Aug 6 at 21:36
Ted Shifrin
59.6k44387
Even with the moving frames computation, you're going to have to do something analogous to the local computation with second-order partial derivatives. How else can we check that a critical point is a local maximum/minimum?
Here's how you should start: Write $dk_i = sumlimits_j k_ijomega_j$ (so we know that $k_ij = 0$ at $M$ for $i,j=1,2$). Then write $dk_ij = sumlimits_ell k_ijellomega_ell$. If $k_1>k_2$ locally, then we know that $k_1jj le 0$ and $k_2jjge 0$ at $M$ for $j=1,2$.
I would rather write your third displayed equations as
beginalign*
dk_1wedgeomega_1 &= (k_2-k_1)omega_12wedgeomega_2 \
dk_2wedgeomega_2 &= (k_1-k_2)omega_1wedgeomega_12.
endalign*
Solve these to obtain $(k_1-k_2)omega_12 = Aomega_1+Bomega_2$. Now can you proceed?
answered Aug 6 at 21:36
Ted Shifrin
59.6k44387
edited Aug 7 at 0:37
answered Aug 6 at 21:36
Ted Shifrin
59.6k44387
answered Aug 6 at 21:36
Ted Shifrin
59.6k44387
answered Aug 6 at 21:36
Ted Shifrin
59.6k44387
59.6k44387
I get $A = k_12$ and $B = k_21$. Then differentiating your last equation I obtain at M $ -(k_1 - k_2)k_1k_2 = -k_122+k_211$, which is a contradiction since the LHS is $lt 0$ and the RHS is $geq 0$. Thank you very very much! I know I had to take the second derivative, but I didn't know how to do it! The "trick" you used - write the differential $dk_i$ with respect of the dual basis and then differentiate the coefficients $k_ij$ instead of the whole 1-form - is great! Thank you again!
– Marco All-in Nervo
Aug 7 at 14:34
Is this the standard way to get second derivatives using differential forms?
– Marco All-in Nervo
Aug 7 at 14:40
Pretty much, yes. It corresponds locally, if you like, to taking the (co)tangent bundle of the (co)tangent bundle.
– Ted Shifrin
Aug 7 at 16:06
Could you expand this a little bit?
– Marco All-in Nervo
Aug 7 at 18:22
You might start with this, but don't get too carried away. If you haven't thought about vector bundles much, I suggest waiting. Intuitively, I just meant that if you think of $1$-form $omega$ as a section of $T^*X$, then the cotangent space to the fiber of $T^*X$ at $omega(p)=sum a_i(p)dx^i(p)$ will consist (locally) of derivatives of $a_i$ at $p$. You can think about local coordinates for $T^*X$, $n=dim X$ coming from $X$ and $n$ coming from the coefficients of a $1$-form with respect to a basis (in our set up, that was the $omega_i$).
– Ted Shifrin
Aug 7 at 18:30
 |Â
show 4 more comments
I get $A = k_12$ and $B = k_21$. Then differentiating your last equation I obtain at M $ -(k_1 - k_2)k_1k_2 = -k_122+k_211$, which is a contradiction since the LHS is $lt 0$ and the RHS is $geq 0$. Thank you very very much! I know I had to take the second derivative, but I didn't know how to do it! The "trick" you used - write the differential $dk_i$ with respect of the dual basis and then differentiate the coefficients $k_ij$ instead of the whole 1-form - is great! Thank you again!
– Marco All-in Nervo
Aug 7 at 14:34
Is this the standard way to get second derivatives using differential forms?
– Marco All-in Nervo
Aug 7 at 14:40
Pretty much, yes. It corresponds locally, if you like, to taking the (co)tangent bundle of the (co)tangent bundle.
– Ted Shifrin
Aug 7 at 16:06
Could you expand this a little bit?
– Marco All-in Nervo
Aug 7 at 18:22
You might start with this, but don't get too carried away. If you haven't thought about vector bundles much, I suggest waiting. Intuitively, I just meant that if you think of $1$-form $omega$ as a section of $T^*X$, then the cotangent space to the fiber of $T^*X$ at $omega(p)=sum a_i(p)dx^i(p)$ will consist (locally) of derivatives of $a_i$ at $p$. You can think about local coordinates for $T^*X$, $n=dim X$ coming from $X$ and $n$ coming from the coefficients of a $1$-form with respect to a basis (in our set up, that was the $omega_i$).
– Ted Shifrin
Aug 7 at 18:30
I get $A = k_12$ and $B = k_21$. Then differentiating your last equation I obtain at M $ -(k_1 - k_2)k_1k_2 = -k_122+k_211$, which is a contradiction since the LHS is $lt 0$ and the RHS is $geq 0$. Thank you very very much! I know I had to take the second derivative, but I didn't know how to do it! The "trick" you used - write the differential $dk_i$ with respect of the dual basis and then differentiate the coefficients $k_ij$ instead of the whole 1-form - is great! Thank you again!
– Marco All-in Nervo
Aug 7 at 14:34
I get $A = k_12$ and $B = k_21$. Then differentiating your last equation I obtain at M $ -(k_1 - k_2)k_1k_2 = -k_122+k_211$, which is a contradiction since the LHS is $lt 0$ and the RHS is $geq 0$. Thank you very very much! I know I had to take the second derivative, but I didn't know how to do it! The "trick" you used - write the differential $dk_i$ with respect of the dual basis and then differentiate the coefficients $k_ij$ instead of the whole 1-form - is great! Thank you again!
– Marco All-in Nervo
Aug 7 at 14:34
Is this the standard way to get second derivatives using differential forms?
– Marco All-in Nervo
Aug 7 at 14:40
Is this the standard way to get second derivatives using differential forms?
– Marco All-in Nervo
Aug 7 at 14:40
Pretty much, yes. It corresponds locally, if you like, to taking the (co)tangent bundle of the (co)tangent bundle.
– Ted Shifrin
Aug 7 at 16:06
Pretty much, yes. It corresponds locally, if you like, to taking the (co)tangent bundle of the (co)tangent bundle.
– Ted Shifrin
Aug 7 at 16:06
Could you expand this a little bit?
– Marco All-in Nervo
Aug 7 at 18:22
Could you expand this a little bit?
– Marco All-in Nervo
Aug 7 at 18:22
You might start with this, but don't get too carried away. If you haven't thought about vector bundles much, I suggest waiting. Intuitively, I just meant that if you think of $1$-form $omega$ as a section of $T^*X$, then the cotangent space to the fiber of $T^*X$ at $omega(p)=sum a_i(p)dx^i(p)$ will consist (locally) of derivatives of $a_i$ at $p$. You can think about local coordinates for $T^*X$, $n=dim X$ coming from $X$ and $n$ coming from the coefficients of a $1$-form with respect to a basis (in our set up, that was the $omega_i$).
– Ted Shifrin
Aug 7 at 18:30
You might start with this, but don't get too carried away. If you haven't thought about vector bundles much, I suggest waiting. Intuitively, I just meant that if you think of $1$-form $omega$ as a section of $T^*X$, then the cotangent space to the fiber of $T^*X$ at $omega(p)=sum a_i(p)dx^i(p)$ will consist (locally) of derivatives of $a_i$ at $p$. You can think about local coordinates for $T^*X$, $n=dim X$ coming from $X$ and $n$ coming from the coefficients of a $1$-form with respect to a basis (in our set up, that was the $omega_i$).
– Ted Shifrin
Aug 7 at 18:30
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So is Cartan being explicit and saying (as in Hilbert's Lemma, which you've quoted from my text) that the smaller principal curvature has a local minimum and the larger principal curvature has a local maximum? Your statement of the question does not include these details.
– Ted Shifrin
Aug 6 at 21:20