3
$\begingroup$

It has been proved that,

If $\lambda_1,\,\lambda_2,\cdots,\lambda_n$ are real analytic functions from $\mathbb{R}^n$ to $\mathbb{R}$, such that $\lambda_i(0)\neq \lambda_j(0)$ for $i\neq j$, then there exists a (real, analytic-) orthonormal coframe $\left(\omega_1,\cdots,\omega_n\right)$ in $(\mathbb{R}^n,g_0)$, with $g_0$ the standard Euclidean metric, defined in a neighbordhood of the origin such that the Riemannian metric

$$ g=\sum_{i=1}^{n}e^{\lambda_i}\omega_i^2 $$

is flat. The condition $\lambda_i(0)\neq \lambda_j(0)$ overwhelms several physical applications, and I wonder if one can prove the same (or a similar) result when two of the functions, say $\lambda_1$ and $\lambda_2$, are identical, or at least when all the $\lambda$'s are constant and two of them are equal. The case $n=3$ is particularly interesting, but a naïve calculation of the Ricci tensor of the metric

$$ g=e^{\lambda_1}\left(\sin\theta\,\mathrm{d} z-\cos\theta\,\left(\cos\phi\,\mathrm{d} x+\sin\phi\,\mathrm{d} y\right)\right)^{2}+e^{\lambda_1}\left(\sin\phi\,\mathrm{d} x-\cos\phi\,\mathrm{d} y\right)^{2}+e^{\lambda_3}\left(\cos\theta\,\mathrm{d} z+\sin\theta\,\left(\cos\phi\,\mathrm{d} x+\sin\phi\,\mathrm{d} y\right)\right)^{2} $$

($\lambda_1\neq\lambda_3$ constanst) explodes, and it's not evident that one can find functions $\theta$ and $\phi$ that render the metric flat.

$\endgroup$
3
  • 3
    $\begingroup$ When all the $\lambda$s are constant, then any "constant" coframe will give you a flat metric. $\endgroup$ Mar 4 at 20:10
  • $\begingroup$ @WillieWong True. How about if we add the requirement of "non-constant" frame. Is there any evident obstruction ? $\endgroup$ Mar 8 at 3:34
  • $\begingroup$ When all the $\lambda$s are constant and two of them are equal (as you wrote), you can also give a non-constant frame by rotating among the equal eigenvalues. For example, suppose you have $\lambda_1 = \lambda_2 \neq \lambda_3$, set $e_i$ the standard coframe and $\omega_1 = \cos(x_3) e_1 + \sin(x_3) e_2$ and $\omega_2 = \sin(x_3) e_1 - \cos(x_3) e_2$ and $\omega_3 = e_3$, then you have a "non-constant" solution. // Rather than trying to impose arbitrary meaningless restrictions, you should just accept that the constant $\lambda$ case is trivial and focus on the case where $\lambda$s vary. $\endgroup$ Mar 8 at 14:11
5
$\begingroup$

Classifying the solutions is a non-trivial problem when two of the $\lambda_i$ are equal, even when they are constants. The reason is that this is essentially an overdetermined problem when two of the $\lambda_i$ are equal.

The point is this: When the $\lambda_i$ are distinct, a choice of an orthonormal coframing $\omega = (\omega_i)$ is essentially a map of $\mathbb{R}^3$ into $\mathrm{SO}(3)$, i.e., a choice of three functions of 3 variables. The requirement that the metric $g$ as defined above be flat is three equations, i.e., the vanishing of the three symmetric functions of the eigenvalues of the Ricci tensor of $g$. It's not surprising that this system of equations has local solutions (when the $\lambda_i$ are distinct and real-analytic). Setting it up as an exterior differential system, one quickly finds that this system is involutive, and the Cartan-Kähler theorem does the job. (N.B.: This fact is not special to $3$-dimensions or to flatness. The Cartan-Kähler analysis I describe in this MO answer generalizes without significant change to cover the case of not-necessarily-constant prescribed real-analytic singular value functions $\lambda_1<\lambda_2<\cdots<\lambda_n$ on $M$ for real-analytic local mappings $f:(M^n,g)\to (N^n, h)$ between any two real-analytic Riemannian manifolds.)

However, when $\lambda_1=\lambda_2\not=\lambda_3$, the situation is entirely different. Now, one doesn't need to specify a complete orthonormal coframe, just $\omega_3$, a unit $1$-form, which is essentially a map of $\mathbb{R}^3$ into $S^2$, i.e., two functions of $3$ variables. Now, if $g_0$ is the Euclidean metric, we can write $g$ in the form $$ g = e^{\lambda_1} g_0 + (e^{\lambda_3}-e^{\lambda_1})\,{\omega_3}^2, $$ and the flatness condition on $g$, i.e., $\mathrm{Ric}(g) = 0$ is still three equations on the two unknowns that go into the definition of $\omega_3$. For generic $\lambda_1\not=\lambda_3$, one does not expect there to be any solutions, even locally, but constructing an actual example of a pair $(\lambda_1,\lambda_3)$ for which no solution exists might be hard. (As a comparison, consider the even more restrictive case in which all the $\lambda_i$ are equal. There, one definitely gets conditions on the function $\lambda = \lambda_1=\lambda_2=\lambda_3$ in order for $e^\lambda g_0$ to be flat. In fact, there is only a 4-parameter family of (local) functions $\lambda$ that satisfy these conditons, and, of course, we know these explicitly.) I expect that there will be some set of PDE relating $\lambda_1$ and $\lambda_3$ whose satisfaction tells whether solutions exist, but I have a feeling that this compatibility condition is liable to be quite complicated.

Even when $\lambda_1$ and $\lambda_3$ are constants, classifying the solutions is not trivial. Clearly, one can reduce to the case in which $\lambda_1=0$ by scaling $g$ by a constant (which won't affect its flatness). For simplicity, set $e^{\lambda_3} = a^2\not=1$. As Willie Wong has observed, taking $\omega_3$ to have constant coefficients always gives a solution, which one can regard as 'trivial'.

Meanwhile, there are nontrivial solutions: Consider the following construction: Let $I\subset\mathbb{R}$ be an interval and let $f = (f_i):I\to S^2$ be a smooth mapping. Consider, the $1$-parameter family of metrics $$h_{f,b}= du^2 + dv^2 + b^2\,\bigl(f_1(t)\,u + f_2(t)\,v + f_3(t)\bigr)^2\,dt^2$$ in the domain $D_f\subset\mathbb{R}^2\times I$ where $h_{f,b}$ is positive definite, i.e., the complement of the locus where $f_1(t)\,u + f_2(t)\,v + f_3(t)=0$. Here $b>0$ is a constant parameter. It is simple to verify that $h_{f,b}$ is a flat metric for any choice of $f$ and $b$. As a result, there is a smooth local isometry $\phi_f:D_f\to\mathbb{R}^3$ so that $\phi_f^*(g_0) = h_{f,1}$, where $g_0$ is the standard metric on $\mathbb{R}^3$. Restricting the domain of $\phi_f$ to some open subset $D'_f\subset D_f$, if necessary, one can arrange that $\phi_f:D'_f\to\mathbb{R}^3$ be injective. Hence there will exist a unique quadratic form $g_{\lambda_3}$ on the (open) image $\phi_f(D'_f)\subset\mathbb{R}^3$ such that $\phi_f^*(g_{\lambda_3}) = h_{f,a}$. By construction, $$ g_{\lambda_3} = g_0 + (e^{\lambda_3}-1)\,{\omega_3}^2, $$ where $\omega_3$ is a $1$-form (of unit size with respect to $g_0$); in fact, $$ \phi_f^*(\omega_3) = \bigl(f_1(t)\,u + f_2(t)\,v + f_3(t)\bigr)\,dt. $$ It's easy to show that this $g_{\lambda_3}$ is a 'trivial' solution if and only if $f_1 = f_2 = 0$, i.e., $f:I\to S^2$ is the constant map to either the north or south pole.

Thus, this construction gives a class of local solutions that essentially depend on two functions of one variable.

The remaining question is whether there are any solutions that are not equivalent to one of these up to isometries in $\mathbb{R}^3$.

What one can say is that, if there are any such solutions for given constants $\lambda_1 = 0$ and $\lambda_3\not=0$, there are not very many, even locally. More precisely, using the Cartan structure equations, one can show the following: Suppose that one has a unit $1$-form $\omega_3$, such that $$ g = g_0 + (e^{\lambda_3}-1)\,{\omega_3}^2 $$ is flat (where $\lambda_3$ is a nonzero constant). Let $X$ be the vector field $g_0$-dual to $\omega_3$ and consider the quadratic form $$ Q = \mathscr{L}_X\bigl(g_0-{\omega_3}^2\bigr). $$ If $Q$ is identically zero, then $g$ arises locally from the above construction using an $f:I\to S^2$. Moreover, there is at most a $6$-dimensional space of local solutions $\omega_3$ that have a non-zero $Q$. (Probably, the space of such solutions has considerably smaller dimension, but I don't know how much smaller.)

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.