1
$\begingroup$

Given a conformal metric $ds^2 = \omega(x_n)(dx_1^2+dx_2^2+\cdots + dx_{n-1}^2+ dx_n^2)$ in $\mathbb R^n$ with conformal factor of one variable $\omega(x_n)$, does there exist an isometric immersion $ds^2 = (dy_1+dy_2+ \cdots)$ of $y_j \rightarrow y_j(x_i)$ in less than $2n-1$ dimensions?

Thanks to "This is much healthier." for helping me with this question.

"This is much healthier." suggested that immersing each variable $x_j$ to be an angle in a polar coordinate system (complex variable) such that $y_j + i y_{j+n-1} = \rho_{j} e^{i x_j}$ would provide a solution to this problem. I continued with this idea and discovered a very simple solution of isometric immersing of the form

\begin{align} ds^2 &=& (dy_1^2 + dy_{1+n-1}^2) + (dy_2^2 + dy_{2+n-1}^2) +\cdots +(dy_{n-1}^2 + dy_{2n-2}^2)+ (dy_{2n-1}^2) \\ ds^2 &=& (d\rho_1^2 + \rho_1^2dx_{1}^2) + (d\rho_2^2 + \rho_2^2dx_{2}^2) +\cdots +(d\rho_{n-1}^2 + \rho_{n-1}^2dx_{n-1}^2)+ (dy_{2n-1}^2) \\ \rho &=& \rho_1 = \rho_2 = \cdots = \rho_{n-1} \\ ds^2 &=& (d\rho^2 + \rho^2dx_{1}^2) + (d\rho^2 + \rho^2dx_{2}^2) +\cdots +(d\rho^2 + \rho^2dx_{n-1}^2)+ (dy_{2n-1}^2) \\ ds^2 &=& \rho^2 \left( (dx_{1}^2) + (dx_{2}^2) + \cdots + (dx_{n-1}^2) + \left( \dfrac{(n-1)d\rho^2 + dy_{2n-1}^2}{\rho^2} \right) \right) \\ ds^2 &=& \omega(x_n)\left((dx_1^2)+(dx_2^2)+\cdots + (dx_{n-1}^2)+ (dx_n^2)\right) \end{align} where $\rho^2 = \omega(x_n)$ and

\begin{align} \dfrac{(n-1)d\rho^2 + dy_{2n-1}^2}{\rho^2} = dx_n^2 \end{align} which leads to the differential equation \begin{align} y_{2n-1}'(x_n) = \sqrt{ (\rho(x_n))^2 - (n-1) (\rho'(x_n))^2 } . \end{align} If one uses the following conformal factor $\omega(x_n) = e^{f(x_n)}$, then \begin{align} y_{2n-1}'(x_n) = e^{f(x_n)} \sqrt{ 1 - (n-1) (f'(x_n))^2 } . \end{align}

This isometric immersion requires $2n-1$ dimensions. Is there away to condense $y_j$ to fewer dimensions?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.