Curvature computation (mixing of partial derivatives and the curvature tensor) [Kähler Geometry]

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Let $(M, omega)$ be a compact Kähler manifold and let $lambda_1$ denote the first eigenvalue of the Laplace operator $Delta$. It is a well known fact that there exists an eigenfunction $u$ such that $Delta u = - lambda_1 u$.

I am working through the following computation on page 65 of Gang Tian's Canonical Metrics in Kähler Geometry text.

At a point, we write the metric as $g_i overlinej = delta_ij$. Then we see that begineqnarray*
u_i j u_overlinei overlinej &=& (u_i j u_overlinei)_overlinej- u_i j overlinej u_overlinei, \
&=& textdiv(u_ij u_overlinei) - u_jioverlinej u_overlinei \
&=& textdiv(u_ij u_overlinei) - u_j overlinej i u_i - R_overlines j i overlinej u_s u_overlinei
endeqnarray*

Here $$u_i overlinej : = fracpartial^2 upartial z_i partial overlinez_j.$$

Can someone explain to me why $u_ij overlinej = u_j i overlinej$, but the mixing of holomorphic and anti-holomorphic partial derivatives picks up a curvature term? Any help with justifying the last line of the computation of $u_ij u_overlinei overlinej$ would be very much appreciated.

Thanks in advance.

pde riemannian-geometry complex-geometry curvature kahler-manifolds

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asked Jul 20 at 22:25

user519758

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Let $(M, omega)$ be a compact Kähler manifold and let $lambda_1$ denote the first eigenvalue of the Laplace operator $Delta$. It is a well known fact that there exists an eigenfunction $u$ such that $Delta u = - lambda_1 u$.

I am working through the following computation on page 65 of Gang Tian's Canonical Metrics in Kähler Geometry text.

At a point, we write the metric as $g_i overlinej = delta_ij$. Then we see that begineqnarray*
u_i j u_overlinei overlinej &=& (u_i j u_overlinei)_overlinej- u_i j overlinej u_overlinei, \
&=& textdiv(u_ij u_overlinei) - u_jioverlinej u_overlinei \
&=& textdiv(u_ij u_overlinei) - u_j overlinej i u_i - R_overlines j i overlinej u_s u_overlinei
endeqnarray*

Here $$u_i overlinej : = fracpartial^2 upartial z_i partial overlinez_j.$$

Can someone explain to me why $u_ij overlinej = u_j i overlinej$, but the mixing of holomorphic and anti-holomorphic partial derivatives picks up a curvature term? Any help with justifying the last line of the computation of $u_ij u_overlinei overlinej$ would be very much appreciated.

Thanks in advance.

pde riemannian-geometry complex-geometry curvature kahler-manifolds

share|cite|improve this question

asked Jul 20 at 22:25

user519758

604

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Let $(M, omega)$ be a compact Kähler manifold and let $lambda_1$ denote the first eigenvalue of the Laplace operator $Delta$. It is a well known fact that there exists an eigenfunction $u$ such that $Delta u = - lambda_1 u$.

I am working through the following computation on page 65 of Gang Tian's Canonical Metrics in Kähler Geometry text.

At a point, we write the metric as $g_i overlinej = delta_ij$. Then we see that begineqnarray*
u_i j u_overlinei overlinej &=& (u_i j u_overlinei)_overlinej- u_i j overlinej u_overlinei, \
&=& textdiv(u_ij u_overlinei) - u_jioverlinej u_overlinei \
&=& textdiv(u_ij u_overlinei) - u_j overlinej i u_i - R_overlines j i overlinej u_s u_overlinei
endeqnarray*

Here $$u_i overlinej : = fracpartial^2 upartial z_i partial overlinez_j.$$

Can someone explain to me why $u_ij overlinej = u_j i overlinej$, but the mixing of holomorphic and anti-holomorphic partial derivatives picks up a curvature term? Any help with justifying the last line of the computation of $u_ij u_overlinei overlinej$ would be very much appreciated.

Thanks in advance.

pde riemannian-geometry complex-geometry curvature kahler-manifolds

share|cite|improve this question

asked Jul 20 at 22:25

user519758

604

Let $(M, omega)$ be a compact Kähler manifold and let $lambda_1$ denote the first eigenvalue of the Laplace operator $Delta$. It is a well known fact that there exists an eigenfunction $u$ such that $Delta u = - lambda_1 u$.

I am working through the following computation on page 65 of Gang Tian's Canonical Metrics in Kähler Geometry text.

At a point, we write the metric as $g_i overlinej = delta_ij$. Then we see that begineqnarray*
u_i j u_overlinei overlinej &=& (u_i j u_overlinei)_overlinej- u_i j overlinej u_overlinei, \
&=& textdiv(u_ij u_overlinei) - u_jioverlinej u_overlinei \
&=& textdiv(u_ij u_overlinei) - u_j overlinej i u_i - R_overlines j i overlinej u_s u_overlinei
endeqnarray*

Here $$u_i overlinej : = fracpartial^2 upartial z_i partial overlinez_j.$$

Can someone explain to me why $u_ij overlinej = u_j i overlinej$, but the mixing of holomorphic and anti-holomorphic partial derivatives picks up a curvature term? Any help with justifying the last line of the computation of $u_ij u_overlinei overlinej$ would be very much appreciated.

Thanks in advance.

pde riemannian-geometry complex-geometry curvature kahler-manifolds

share|cite|improve this question

asked Jul 20 at 22:25

user519758

604

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 22:25

user519758

604

asked Jul 20 at 22:25

user519758

604

asked Jul 20 at 22:25

user519758

604

604

add a comment | 

add a comment | 

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In Riemannian manifold, $nabla_Znabla_Xnabla_Y f-nabla_Z
nabla_Ynabla _Xf=0$ so that we have it.

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answered Jul 27 at 4:13

HK Lee

13.5k31855

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1 Answer
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active

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votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

In Riemannian manifold, $nabla_Znabla_Xnabla_Y f-nabla_Z
nabla_Ynabla _Xf=0$ so that we have it.

share|cite|improve this answer

answered Jul 27 at 4:13

HK Lee

13.5k31855

add a comment | 

up vote
0
down vote

In Riemannian manifold, $nabla_Znabla_Xnabla_Y f-nabla_Z
nabla_Ynabla _Xf=0$ so that we have it.

share|cite|improve this answer

answered Jul 27 at 4:13

HK Lee

13.5k31855

add a comment | 

up vote
0
down vote

up vote
0
down vote

In Riemannian manifold, $nabla_Znabla_Xnabla_Y f-nabla_Z
nabla_Ynabla _Xf=0$ so that we have it.

share|cite|improve this answer

answered Jul 27 at 4:13

HK Lee

13.5k31855

In Riemannian manifold, $nabla_Znabla_Xnabla_Y f-nabla_Z
nabla_Ynabla _Xf=0$ so that we have it.

share|cite|improve this answer

answered Jul 27 at 4:13

HK Lee

13.5k31855

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 4:13

HK Lee

13.5k31855

answered Jul 27 at 4:13

HK Lee

13.5k31855

answered Jul 27 at 4:13

HK Lee

13.5k31855

13.5k31855

add a comment | 

add a comment | 

 

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