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- Summary
- Thread involves finding the dimension of a tangent subspace ...

I am reading Andrew McInerney's book: First Steps in Differential Geometry: Riemannian, Contact, Symplectic ... and I am focused on Chapter 3: Advanced Calculus ... and in particular on Section 3.3: Geometric Sets and Subspaces of ##T_p ( \mathbb{R}^n )## ... ...

I need help with an aspect of the proof of Theorem 3.3.13 ... ...

Theorem 3.3.13 (together with a relevant definition) reads as follows:

In the above text from McInerney we read the following:

" ... ... The fact that ##\phi## has rank ##n -1## follows by computing the Jacobian matrix at any point in ##U## ... ... "

Can someone please demonstrate rigorously, formally and explicitly that ##\phi## has rank ##n -1## ... ...

My computations with respect to the Jacobian ##[D \phi(p) ]## were as follows:

We have ##\phi ( x_1, \ ... \ ... \ x_{n-1} ) = ( x_1, \ ... \ ... \ x_{n-1}, f( x_1, \ ... \ ... \ x_{n-1}) )##

Now put ...

##f_1( x_1, \ ... \ ... \ x_{n-1} ) = x_1##

##f_2( x_1, \ ... \ ... \ x_{n-1} ) = x_2##

... ... ...

... ... ...

##f_{n-1}( x_1, \ ... \ ... \ x_{n-1} ) = x_{n-1}##

##f_n( x_1, \ ... \ ... \ x_{n-1} ) = f( x_1, \ ... \ ... \ x_{n-1} )##

Then ... the Jacobian ...

##[D \phi(p) ] = \begin{bmatrix} \frac{ \partial f_1 }{ \partial x_1} & ... & ... & \frac{ \partial f_1 }{ \partial x_{n-1} } \\ ... & ... & ... & ... \\ ... & ... & ... & ... \\ \frac{ \partial f_{n-1} }{ \partial x_1} & ... & ... & \frac{ \partial f_{n-1} }{ \partial x_{n-1} } \\ \frac{ \partial f }{ \partial x_1} & ... & ... & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}##

##= \begin{bmatrix} 1 & 0 & 0 & ... & ... & 0 \\ 0 & 1 & 0 & ... & ... & 0 \\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... & ... \\ \frac{ \partial f }{ \partial x_1} & ... & ... & ... & ... & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}##

... now ... how do we show that the rank of ##[D \phi(p) ]## is ##n-1## ...?

Hope someone can help ...

Peter

I need help with an aspect of the proof of Theorem 3.3.13 ... ...

Theorem 3.3.13 (together with a relevant definition) reads as follows:

In the above text from McInerney we read the following:

" ... ... The fact that ##\phi## has rank ##n -1## follows by computing the Jacobian matrix at any point in ##U## ... ... "

Can someone please demonstrate rigorously, formally and explicitly that ##\phi## has rank ##n -1## ... ...

My computations with respect to the Jacobian ##[D \phi(p) ]## were as follows:

We have ##\phi ( x_1, \ ... \ ... \ x_{n-1} ) = ( x_1, \ ... \ ... \ x_{n-1}, f( x_1, \ ... \ ... \ x_{n-1}) )##

Now put ...

##f_1( x_1, \ ... \ ... \ x_{n-1} ) = x_1##

##f_2( x_1, \ ... \ ... \ x_{n-1} ) = x_2##

... ... ...

... ... ...

##f_{n-1}( x_1, \ ... \ ... \ x_{n-1} ) = x_{n-1}##

##f_n( x_1, \ ... \ ... \ x_{n-1} ) = f( x_1, \ ... \ ... \ x_{n-1} )##

Then ... the Jacobian ...

##[D \phi(p) ] = \begin{bmatrix} \frac{ \partial f_1 }{ \partial x_1} & ... & ... & \frac{ \partial f_1 }{ \partial x_{n-1} } \\ ... & ... & ... & ... \\ ... & ... & ... & ... \\ \frac{ \partial f_{n-1} }{ \partial x_1} & ... & ... & \frac{ \partial f_{n-1} }{ \partial x_{n-1} } \\ \frac{ \partial f }{ \partial x_1} & ... & ... & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}##

##= \begin{bmatrix} 1 & 0 & 0 & ... & ... & 0 \\ 0 & 1 & 0 & ... & ... & 0 \\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... & ... \\ \frac{ \partial f }{ \partial x_1} & ... & ... & ... & ... & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}##

... now ... how do we show that the rank of ##[D \phi(p) ]## is ##n-1## ...?

Hope someone can help ...

Peter