What is the mistake in my coordinate transformation for Theorema Egregium?

[Simone Furcas]

I have ##ds^2=\cos^2(v)du^2 + dv^2## , i take a coordinate transformation x=u and cos(v)=##\frac{1}{(cosh(y))}##, I have to find a new metric with this coordinate transformation and proof it is in agreement with Theorema Egregium. ##ds^2=\frac{dx^2}{(cosh^2(y)}) +\frac{dy^2 }{(y^2(1-y^2))}## is my new metric . I used the jacobian to proof the second point, ##J=\begin{pmatrix} 1 & 0 \\ 0 & \frac{-1}{y(1-y^2)^(1/2)} \end{pmatrix}##, so ##j^-1 * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{y^2(1-y^2)} \end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{y^2(1-y^2)})^2 \end{pmatrix}##
I know cos(v)=##\frac{1}{(cosh(y))}## so the first part is ok, the second ##(\frac{1}{(y^2(1-y^2)})^2##=1 i think is not true...
What is my mistake? Does someone could help me?

 

[Simone Furcas]

I have ##ds^2=\cos^2(v)du^2 + dv^2## , i take a coordinate transformation x=u and cos(v)=##\frac{1}{(cosh(y))}##, I have to find a new metric with this coordinate transformation and proof it is in agreement with Theorema Egregium. ##ds^2=\frac{dx^2}{(cosh^2(y)}) +\frac{dy^2 }{(y^2(1-y^2))}## is my new metric . I used the jacobian to proof the second point, ##J=\begin{pmatrix} 1 & 0 \\ 0 & \frac{-1}{y(1-y^2)^(1/2)} \end{pmatrix}##, so ##j^-1 * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{y^2(1-y^2)} \end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{y^2(1-y^2)})^2 \end{pmatrix}##
I know cos(v)=##\frac{1}{(cosh(y))}## so the first part is ok, the second ##(\frac{1}{(y^2(1-y^2)})^2##=1 i think is not true...
What is my mistake? Does someone could help me?

I did a mistake, ##ds^2=\frac{dx^2}{(cosh^2(y)}) +\frac{dy^2 }{cosh{y}}##, ##J=\begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{cosh{y}} \end{pmatrix}##, so ##j^-1 * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{cosh{y}}\end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{cosh{y}})^4 \end{pmatrix}##
Does someone have any ideas?

 

[HallsofIvy]

That matrix multiplication is wrong. Since these are diagonal matrices multiplication is commutative so [itex]J^{-1}MJ= (J^{-1}J)M= M[/itex] for any matrix, M.

 

[Simone Furcas]

I did a mistake, ##ds^2=\frac{dx^2}{(cosh^2(y)}) +\frac{dy^2 }{cosh{y}}##, ##J=\begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{cosh{y}} \end{pmatrix}##, so ##j^-1 * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{cosh{y}}\end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{cosh{y}})^4 \end{pmatrix}##
Does someone have any ideas?

##j^T * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{cosh{y}}\end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{cosh{y}})^4 \end{pmatrix}## there was a mistake, it was the transpose not the inverse.

 

[Simone Furcas]

I solved it.

##j^T * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{cosh{y}}\end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{cosh{y}})^4 \end{pmatrix}## there was a mistake, it was the transpose not the inverse.

I solved the exercise.