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Isometric Immersions: equivalent condition for $R^perp=0$

Totally geodesic immersionsDoes every smooth manifold admits non-isometric riemannian metrics?Two successive isometric immersions: relation between mean curvature vectors?Does harmonic decomposition preserve immersions?Eigenvalues of shape operator and of curvature on second exterior powerIsometric immersions between manifolds with boundary are locally distance preserving?Equivalent condition for $f:Mto N$ to be an embeddingNon equivalent isometric embeddings of the planeName of the relation between LC connections in isometric immersionsConfused about Isometric Immersions

0

$begingroup$

In Dajczer's Submanifolds and Isometric Immersions, a couple paragraphs before presenting the Fundamental Theorem for Submanifolds, the author presents Ricci's equation:
$$(widetildeR(X,Y)xi)^perp=R^perp(X,Y)xi+alpha(A_xi X,Y)-alpha(X,A_xi Y)text, forall X,Yin T_xM, xiin T_xM^perp$$

(where $alpha$ is second fundamental form and $A$ is the shape operator)

He points out that if $widetildeM$ has constant sectional curvature, Ricci's formula becomes:
$$R^perp(X,Y)xi=alpha(X,A_xi Y)-alpha(A_xi X, Y)$$

Then he says: "notice that $R^perp_x=0$ if and only if there exists an orthogonal basis for $T_xM$ that diagonalizes simultaneously all $A_xi, xiin T_xM^perp$"

If I understand correctly, he means there is a basis $v_1,...,v_n$ of $T_xM$ with $langle v_i,v_jrangle=delta_ij$ and such that $A_xi v_i=lambda_iv_i$ for all $i$ and for all $xiin T_xM^perp$.

The way he mentions it makes it look like something obvious, but I really can't see how this relates to $R_x^perp=0$. Am I missing something?

riemannian-geometry smooth-manifolds connections

share|cite|improve this question

asked Mar 17 at 23:17

rmdmc89rmdmc89

2,2651923

$endgroup$

add a comment | 

0

$begingroup$

In Dajczer's Submanifolds and Isometric Immersions, a couple paragraphs before presenting the Fundamental Theorem for Submanifolds, the author presents Ricci's equation:
$$(widetildeR(X,Y)xi)^perp=R^perp(X,Y)xi+alpha(A_xi X,Y)-alpha(X,A_xi Y)text, forall X,Yin T_xM, xiin T_xM^perp$$

(where $alpha$ is second fundamental form and $A$ is the shape operator)

He points out that if $widetildeM$ has constant sectional curvature, Ricci's formula becomes:
$$R^perp(X,Y)xi=alpha(X,A_xi Y)-alpha(A_xi X, Y)$$

Then he says: "notice that $R^perp_x=0$ if and only if there exists an orthogonal basis for $T_xM$ that diagonalizes simultaneously all $A_xi, xiin T_xM^perp$"

If I understand correctly, he means there is a basis $v_1,...,v_n$ of $T_xM$ with $langle v_i,v_jrangle=delta_ij$ and such that $A_xi v_i=lambda_iv_i$ for all $i$ and for all $xiin T_xM^perp$.

The way he mentions it makes it look like something obvious, but I really can't see how this relates to $R_x^perp=0$. Am I missing something?

riemannian-geometry smooth-manifolds connections

share|cite|improve this question

asked Mar 17 at 23:17

rmdmc89rmdmc89

2,2651923

$endgroup$

add a comment | 

$begingroup$

In Dajczer's Submanifolds and Isometric Immersions, a couple paragraphs before presenting the Fundamental Theorem for Submanifolds, the author presents Ricci's equation:
$$(widetildeR(X,Y)xi)^perp=R^perp(X,Y)xi+alpha(A_xi X,Y)-alpha(X,A_xi Y)text, forall X,Yin T_xM, xiin T_xM^perp$$

(where $alpha$ is second fundamental form and $A$ is the shape operator)

He points out that if $widetildeM$ has constant sectional curvature, Ricci's formula becomes:
$$R^perp(X,Y)xi=alpha(X,A_xi Y)-alpha(A_xi X, Y)$$

Then he says: "notice that $R^perp_x=0$ if and only if there exists an orthogonal basis for $T_xM$ that diagonalizes simultaneously all $A_xi, xiin T_xM^perp$"

If I understand correctly, he means there is a basis $v_1,...,v_n$ of $T_xM$ with $langle v_i,v_jrangle=delta_ij$ and such that $A_xi v_i=lambda_iv_i$ for all $i$ and for all $xiin T_xM^perp$.

The way he mentions it makes it look like something obvious, but I really can't see how this relates to $R_x^perp=0$. Am I missing something?

riemannian-geometry smooth-manifolds connections

share|cite|improve this question

asked Mar 17 at 23:17

rmdmc89rmdmc89

2,2651923

$endgroup$

share|cite|improve this question

asked Mar 17 at 23:17

rmdmc89rmdmc89

2,2651923

share|cite|improve this question

asked Mar 17 at 23:17

rmdmc89rmdmc89

2,2651923

asked Mar 17 at 23:17

rmdmc89rmdmc89

2,2651923

asked Mar 17 at 23:17

rmdmc89rmdmc89

2,2651923

1 Answer
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1

$begingroup$

Take the inner product of the Ricci formula with another normal vector $etain T_x M^perp$. You get
$$
beginalign*
langle R^perp(X,Y)xi,etarangle &= langle alpha(X,A_xi Y),eta rangle
- langle alpha(A_xi X,Y),eta rangle \
&= langle A_eta X, A_xi Y rangle
- langle A_eta A_xi X, Y rangle \
&= langle A_xi A_eta X, Y rangle
- langle A_eta A_xi X, Y rangle \
&= langle [A_xi, A_eta]X, Yrangle.
endalign*
$$
Now you see that $R^perp_x$ vanishes iff the commutators $[A_xi, A_eta]$ of the shape operators vanish for all $xi, eta in T_x^perp M$. Recall that the shape operators are diagonalisable, since they are symmetric. It is a well-known fact that diagonalisable linear maps can be diagonalised simultaneously if and only if they commute. This shows the statement.

share|cite|improve this answer

edited Mar 18 at 9:36

answered Mar 17 at 23:30

Ernie060Ernie060

2,795619

$endgroup$

add a comment | 

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1

$begingroup$

Take the inner product of the Ricci formula with another normal vector $etain T_x M^perp$. You get
$$
beginalign*
langle R^perp(X,Y)xi,etarangle &= langle alpha(X,A_xi Y),eta rangle
- langle alpha(A_xi X,Y),eta rangle \
&= langle A_eta X, A_xi Y rangle
- langle A_eta A_xi X, Y rangle \
&= langle A_xi A_eta X, Y rangle
- langle A_eta A_xi X, Y rangle \
&= langle [A_xi, A_eta]X, Yrangle.
endalign*
$$
Now you see that $R^perp_x$ vanishes iff the commutators $[A_xi, A_eta]$ of the shape operators vanish for all $xi, eta in T_x^perp M$. Recall that the shape operators are diagonalisable, since they are symmetric. It is a well-known fact that diagonalisable linear maps can be diagonalised simultaneously if and only if they commute. This shows the statement.

share|cite|improve this answer

edited Mar 18 at 9:36

answered Mar 17 at 23:30

Ernie060Ernie060

2,795619

$endgroup$

add a comment | 

1

$begingroup$

Take the inner product of the Ricci formula with another normal vector $etain T_x M^perp$. You get
$$
beginalign*
langle R^perp(X,Y)xi,etarangle &= langle alpha(X,A_xi Y),eta rangle
- langle alpha(A_xi X,Y),eta rangle \
&= langle A_eta X, A_xi Y rangle
- langle A_eta A_xi X, Y rangle \
&= langle A_xi A_eta X, Y rangle
- langle A_eta A_xi X, Y rangle \
&= langle [A_xi, A_eta]X, Yrangle.
endalign*
$$
Now you see that $R^perp_x$ vanishes iff the commutators $[A_xi, A_eta]$ of the shape operators vanish for all $xi, eta in T_x^perp M$. Recall that the shape operators are diagonalisable, since they are symmetric. It is a well-known fact that diagonalisable linear maps can be diagonalised simultaneously if and only if they commute. This shows the statement.

share|cite|improve this answer

edited Mar 18 at 9:36

answered Mar 17 at 23:30

Ernie060Ernie060

2,795619

$endgroup$

add a comment | 

$begingroup$

Take the inner product of the Ricci formula with another normal vector $etain T_x M^perp$. You get
$$
beginalign*
langle R^perp(X,Y)xi,etarangle &= langle alpha(X,A_xi Y),eta rangle
- langle alpha(A_xi X,Y),eta rangle \
&= langle A_eta X, A_xi Y rangle
- langle A_eta A_xi X, Y rangle \
&= langle A_xi A_eta X, Y rangle
- langle A_eta A_xi X, Y rangle \
&= langle [A_xi, A_eta]X, Yrangle.
endalign*
$$
Now you see that $R^perp_x$ vanishes iff the commutators $[A_xi, A_eta]$ of the shape operators vanish for all $xi, eta in T_x^perp M$. Recall that the shape operators are diagonalisable, since they are symmetric. It is a well-known fact that diagonalisable linear maps can be diagonalised simultaneously if and only if they commute. This shows the statement.

share|cite|improve this answer

edited Mar 18 at 9:36

answered Mar 17 at 23:30

Ernie060Ernie060

2,795619

$endgroup$

share|cite|improve this answer

edited Mar 18 at 9:36

answered Mar 17 at 23:30

Ernie060Ernie060

2,795619

answered Mar 17 at 23:30

Ernie060Ernie060

2,795619

answered Mar 17 at 23:30

Ernie060Ernie060

2,795619