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Tangent Spaces and Subspaces ... McInerney Theorem 3.3.13 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
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 \(\displaystyle T_p ( \mathbb{R}^n )\) ... ...

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:


McInerney - Defn 3.3.12 & Theorem 3.3.13 ... ... Page 1 ... .png
McInerney - 2 - Defn 3.3.12 & Theorem 3.3.13 ... ... Page 2 ... .png



In the above text from McInerney we read the following:

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


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


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

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

Now put ...

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

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

... ... ...

... ... ...

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

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


Then ... the Jacobian ...


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


= \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 \(\displaystyle [D \phi(p) ]\) is \(\displaystyle n-1\) ...?




Hope someone can help ...

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,655
Leeds, UK
$[D \phi(p) ] = \begin{bmatrix} 1 & 0 & \ldots & \ldots & 0 \\ 0 & 1 & \ldots & \ldots & 0 \\ \ldots & \ldots & \ddots & \ldots & 0 \\ \ldots & \ldots & \ldots & \ddots & 0 \\ 0&0&\ldots&\ldots&1 \\ \frac{ \partial f_n }{ \partial x_1} & \ldots & \ldots & \ldots & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}$ is an $n\times(n-1)$ matrix, so it cannot have rank greater than $n-1$. On the other hand, its first $n-1$ rows form the identity $(n-1)\times(n-1)$ matrix, which has rank $n-1$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
$[D \phi(p) ] = \begin{bmatrix} 1 & 0 & \ldots & \ldots & 0 \\ 0 & 1 & \ldots & \ldots & 0 \\ \ldots & \ldots & \ddots & \ldots & 0 \\ \ldots & \ldots & \ldots & \ddots & 0 \\ 0&0&\ldots&\ldots&1 \\ \frac{ \partial f_n }{ \partial x_1} & \ldots & \ldots & \ldots & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}$ is an $n\times(n-1)$ matrix, so it cannot have rank greater than $n-1$. On the other hand, its first $n-1$ rows form the identity $(n-1)\times(n-1)$ matrix, which has rank $n-1$.

Thanks Opalg ...

Appreciate your help ...

Peter