Computing $int_{S^{1} times S^{1}} dtheta_{1} wedge dtheta_{2}$
$begingroup$
Let $S^{1} times S^{1}$ be the torus embedded in $mathbb{R}^{4}$. I want to compute
$int_{S^{1} times S^{1}} dtheta_{1} wedge dtheta_{2}$
I believe this should be "essentially" $int_{0}^{2pi} int_{0}^{2pi} dtheta_{1} dtheta_{2}= 4 pi^{2}$. I mainly just wanted to see justification.
I've been reading about differential forms in Guillemin and Pollack, where I did a similar exercise, where I saw that for $$omega = frac{x dy - y dx}{x^2+y^2}$$, the pull-back allows under the map $h(t)=(cos(t), sin(t))$ allows us to see that (see here):
$$int_{S^{1}} omega = int^{2pi}_{0} h^{*}omega=2pi$$
I feel like in this case, we have something similar with $$omega_{i} = frac{x_{i} dy_{i} - y_{i} dx_{i}}{x_{i}^2+y_{i}^2}$$
and considering $$int_{S^{1} times S^{1}} omega_{1} wedge omega_{2}$$
in similar fashion change coordinates say by considering a map $H:(s,t) mapsto (h(s), h(t))$ and considering the pull-back. Would this be a way to justify it?
differential-topology differential-forms
asked Dec 11 '18 at 2:16
user135520user135520
943718
$endgroup$
add a comment |
$begingroup$
Let $S^{1} times S^{1}$ be the torus embedded in $mathbb{R}^{4}$. I want to compute
$int_{S^{1} times S^{1}} dtheta_{1} wedge dtheta_{2}$
I believe this should be "essentially" $int_{0}^{2pi} int_{0}^{2pi} dtheta_{1} dtheta_{2}= 4 pi^{2}$. I mainly just wanted to see justification.
I've been reading about differential forms in Guillemin and Pollack, where I did a similar exercise, where I saw that for $$omega = frac{x dy - y dx}{x^2+y^2}$$, the pull-back allows under the map $h(t)=(cos(t), sin(t))$ allows us to see that (see here):
$$int_{S^{1}} omega = int^{2pi}_{0} h^{*}omega=2pi$$
I feel like in this case, we have something similar with $$omega_{i} = frac{x_{i} dy_{i} - y_{i} dx_{i}}{x_{i}^2+y_{i}^2}$$
and considering $$int_{S^{1} times S^{1}} omega_{1} wedge omega_{2}$$
in similar fashion change coordinates say by considering a map $H:(s,t) mapsto (h(s), h(t))$ and considering the pull-back. Would this be a way to justify it?
differential-topology differential-forms
asked Dec 11 '18 at 2:16
user135520user135520
943718
$endgroup$
$begingroup$
Read the definition of the integral carefully and proceed step by step.
$endgroup$
– Charlie Frohman
Dec 11 '18 at 3:06
2
$begingroup$
@CharlieFrohman, If they could do that then why would they be asking?
$endgroup$
– Chickenmancer
Dec 11 '18 at 16:50
1
$begingroup$
Yes, you're proceeding correctly. All you need at the end is Fubini's Theorem, to turn the double integral into the product of single integrals.
$endgroup$
– Ted Shifrin
Dec 13 '18 at 17:52
$begingroup$
Thank you! So we pull back via $H$ to get $int_{(0, 2pi) times (0, 2pi)} d{theta_{1}} wedge dtheta_{2}$ and then apply Fubini's theorem?
$endgroup$
– user135520
Dec 14 '18 at 16:06
add a comment |
$begingroup$
Let $S^{1} times S^{1}$ be the torus embedded in $mathbb{R}^{4}$. I want to compute
$int_{S^{1} times S^{1}} dtheta_{1} wedge dtheta_{2}$
I believe this should be "essentially" $int_{0}^{2pi} int_{0}^{2pi} dtheta_{1} dtheta_{2}= 4 pi^{2}$. I mainly just wanted to see justification.
I've been reading about differential forms in Guillemin and Pollack, where I did a similar exercise, where I saw that for $$omega = frac{x dy - y dx}{x^2+y^2}$$, the pull-back allows under the map $h(t)=(cos(t), sin(t))$ allows us to see that (see here):
$$int_{S^{1}} omega = int^{2pi}_{0} h^{*}omega=2pi$$
I feel like in this case, we have something similar with $$omega_{i} = frac{x_{i} dy_{i} - y_{i} dx_{i}}{x_{i}^2+y_{i}^2}$$
and considering $$int_{S^{1} times S^{1}} omega_{1} wedge omega_{2}$$
in similar fashion change coordinates say by considering a map $H:(s,t) mapsto (h(s), h(t))$ and considering the pull-back. Would this be a way to justify it?
differential-topology differential-forms
asked Dec 11 '18 at 2:16
user135520user135520
943718
$endgroup$
Let $S^{1} times S^{1}$ be the torus embedded in $mathbb{R}^{4}$. I want to compute
$int_{S^{1} times S^{1}} dtheta_{1} wedge dtheta_{2}$
I believe this should be "essentially" $int_{0}^{2pi} int_{0}^{2pi} dtheta_{1} dtheta_{2}= 4 pi^{2}$. I mainly just wanted to see justification.
I've been reading about differential forms in Guillemin and Pollack, where I did a similar exercise, where I saw that for $$omega = frac{x dy - y dx}{x^2+y^2}$$, the pull-back allows under the map $h(t)=(cos(t), sin(t))$ allows us to see that (see here):
$$int_{S^{1}} omega = int^{2pi}_{0} h^{*}omega=2pi$$
I feel like in this case, we have something similar with $$omega_{i} = frac{x_{i} dy_{i} - y_{i} dx_{i}}{x_{i}^2+y_{i}^2}$$
and considering $$int_{S^{1} times S^{1}} omega_{1} wedge omega_{2}$$
in similar fashion change coordinates say by considering a map $H:(s,t) mapsto (h(s), h(t))$ and considering the pull-back. Would this be a way to justify it?
differential-topology differential-forms
differential-topology differential-forms
asked Dec 11 '18 at 2:16
user135520user135520
943718
asked Dec 11 '18 at 2:16
user135520user135520
943718
asked Dec 11 '18 at 2:16
user135520user135520
943718
asked Dec 11 '18 at 2:16
user135520user135520
943718
asked Dec 11 '18 at 2:16
user135520user135520
943718
943718
$begingroup$
Read the definition of the integral carefully and proceed step by step.
$endgroup$
– Charlie Frohman
Dec 11 '18 at 3:06
2
$begingroup$
@CharlieFrohman, If they could do that then why would they be asking?
$endgroup$
– Chickenmancer
Dec 11 '18 at 16:50
1
$begingroup$
Yes, you're proceeding correctly. All you need at the end is Fubini's Theorem, to turn the double integral into the product of single integrals.
$endgroup$
– Ted Shifrin
Dec 13 '18 at 17:52
$begingroup$
Thank you! So we pull back via $H$ to get $int_{(0, 2pi) times (0, 2pi)} d{theta_{1}} wedge dtheta_{2}$ and then apply Fubini's theorem?
$endgroup$
– user135520
Dec 14 '18 at 16:06
add a comment |
$begingroup$
Read the definition of the integral carefully and proceed step by step.
$endgroup$
– Charlie Frohman
Dec 11 '18 at 3:06
2
$begingroup$
@CharlieFrohman, If they could do that then why would they be asking?
$endgroup$
– Chickenmancer
Dec 11 '18 at 16:50
1
$begingroup$
Yes, you're proceeding correctly. All you need at the end is Fubini's Theorem, to turn the double integral into the product of single integrals.
$endgroup$
– Ted Shifrin
Dec 13 '18 at 17:52
$begingroup$
Thank you! So we pull back via $H$ to get $int_{(0, 2pi) times (0, 2pi)} d{theta_{1}} wedge dtheta_{2}$ and then apply Fubini's theorem?
$endgroup$
– user135520
Dec 14 '18 at 16:06
$begingroup$
Read the definition of the integral carefully and proceed step by step.
$endgroup$
– Charlie Frohman
Dec 11 '18 at 3:06
$begingroup$
Read the definition of the integral carefully and proceed step by step.
$endgroup$
– Charlie Frohman
Dec 11 '18 at 3:06
2
2
$begingroup$
@CharlieFrohman, If they could do that then why would they be asking?
$endgroup$
– Chickenmancer
Dec 11 '18 at 16:50
$begingroup$
@CharlieFrohman, If they could do that then why would they be asking?
$endgroup$
– Chickenmancer
Dec 11 '18 at 16:50
1
1
$begingroup$
Yes, you're proceeding correctly. All you need at the end is Fubini's Theorem, to turn the double integral into the product of single integrals.
$endgroup$
– Ted Shifrin
Dec 13 '18 at 17:52
$begingroup$
Yes, you're proceeding correctly. All you need at the end is Fubini's Theorem, to turn the double integral into the product of single integrals.
$endgroup$
– Ted Shifrin
Dec 13 '18 at 17:52
$begingroup$
Thank you! So we pull back via $H$ to get $int_{(0, 2pi) times (0, 2pi)} d{theta_{1}} wedge dtheta_{2}$ and then apply Fubini's theorem?
$endgroup$
– user135520
Dec 14 '18 at 16:06
$begingroup$
Thank you! So we pull back via $H$ to get $int_{(0, 2pi) times (0, 2pi)} d{theta_{1}} wedge dtheta_{2}$ and then apply Fubini's theorem?
$endgroup$
– user135520
Dec 14 '18 at 16:06
add a comment |
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$begingroup$
Read the definition of the integral carefully and proceed step by step.
$endgroup$
– Charlie Frohman
Dec 11 '18 at 3:06
2
$begingroup$
@CharlieFrohman, If they could do that then why would they be asking?
$endgroup$
– Chickenmancer
Dec 11 '18 at 16:50
1
$begingroup$
Yes, you're proceeding correctly. All you need at the end is Fubini's Theorem, to turn the double integral into the product of single integrals.
$endgroup$
– Ted Shifrin
Dec 13 '18 at 17:52
$begingroup$
Thank you! So we pull back via $H$ to get $int_{(0, 2pi) times (0, 2pi)} d{theta_{1}} wedge dtheta_{2}$ and then apply Fubini's theorem?
$endgroup$
– user135520
Dec 14 '18 at 16:06