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In Pressley book I found problem: Find the first fundamental form and determine what kind of surface is patch $\sigma=\langle u-v,u+v,u^2+v^2\rangle$. It is easy to see that the first fundamental form is $(2+4u^2)du^2+8uv\,du\,dv+(2+4v^2)dv^2$. The answer from the book tells that the surface is paraboloid of revolution. But the surface of revolution of unit-speed curve $\gamma(u)$ around $z$-axis is $\sigma(u,v)=\langle u\cos v, u\sin v,\gamma(u)\rangle$ where $v$ is the angle of rotation.Then the first fundamental form for the surface of rotation is $2du^2+u^2dv^2$, which differs from the result above. Is there an algorithm how to assign unknown surfaces to the typical surfaces relying on the first fundamental form?

pabodu
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A surface $S$ admits many parametrizations/patches $\sigma : D \subset \mathbb{R}^{2}_{(u, v)} \to \mathbb{R}^{3}_{(x,y,z)}$ and in each paramatrization the first fundamental form will potentially take on a different appearance. (Note that this exhibits one of the main features of differential geometry: you are trying to study surfaces independently of the parametrization and must pay attention to how expressions/quantities change when one changes parametrizations (or coordinates).)

In particular, if your surface $S$ is a surface of revolution obtained by revolving a (unit-speed) profile curve $\gamma(u) = (f(u), 0, g(u))$ about the $z$-axis by an angle of $v$ then your parametrization/patch will be of the form $$ \sigma(u, v) = \left\langle f(u)\cos v , f(u) \sin v, g(u)\right\rangle, $$ with corresponding first fundamental form $$ \mathrm{d}s^2 = \mathrm{d}u^2 + f(u)^2\mathrm{d}v^2. $$

But the given parametrization of your surface $S$ is $$ \sigma(u, v) = \left\langle u - v , u + v, u^2 + v^2\right\rangle $$ is not the parametrization obtained by rotating a profile curve about an axis of revolution.

In this instance, the author probably intends that you recognize the surface $S$ as a paraboloid of revolution by noting the following algebraic relation of the component/coordinate functions of the parametrization: \begin{align*} x^2 + y^2 &= x(u, v)^2 + y(u, v)^2\\ &= (u - v)^2 + (u + v)^2\\ &= 2(u^2 + v^2)\\ &= 2z(u, v)\\ &= 2z.\\ \end{align*}

Thus, all points $P(x, y, z)$ on the surface patch satisfy $x^2 + y^2 = 2z$ and lie on the paraboloid of revolution given by the graph of the function $z = F(x, y) = \frac{1}{2}\left(x^2 + y^2\right)$. (Recognizing this would also allow you to know parametrize your surface using Monge patch, in which case the first fundamental form would again change appearance.)

THW
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    The direct substitution does provide the paraboloid. I hoped that it is possible to understand shapes relying on the 1st fundamental form only. You have explained: that's impossible. The negative answer is also an answer. Thank you very much! – pabodu Oct 24 '18 at 17:41
  • However, the next question raises: since the first fundamental form for rotating curve $\langle f(u),0,g(u)\rangle$ is $ds^2=\dot{f}^2du^2+f^2dv^2$, can we guarantee that all surfaces with that fundamental form are the surfaces of rotation? Are there any non-rotational surfaces with the same form $ds^2=\dot{f}^2du^2+f^2dv^2$ ? – pabodu Oct 24 '18 at 18:05
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    Just a small arithmetic mistake that doesn't change the answer qualitatively: In the last display, we should have $(u - v)^2 + (u + v)^2 = 2 (u^2 + v^2)$. – Travis Willse Oct 24 '18 at 19:42
  • Thanks, Travis. Typos should now be fixed. – THW Oct 24 '18 at 20:05
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    @pabodu If you have a surface with first fundamental form $ds^2 = \dot{f}^2\mathrm{d}u^2 + f^2 \mathrm{d}v^2$, then the surface admits a one parameter family of isometries $\psi_{t} : S \to S$ defined defined by $\psi_{t}(u, v) = (\bar{u}, \bar{v}) = (u, v + t)$. The one parameter family of isometries is a property shared with surfaces of revolution, but this property would be shared by any surface with a first fundamental form of the form $\mathrm{d}s^2 = A(u)\mathrm{d}u^2 + B(u)\mathrm{d}v^2$, $A(u), B(u) > 0$. (The same one parameter family of isometries would work again.) cont. . . – THW Oct 25 '18 at 00:36
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    . . . but the one parameter family of isometries isn't going to be enough. For example, you could have a surface like Enneper's surface, where the first fundamental form in the standard parametrization is $\mathrm{d}s^2 = (1 + u^2 + v^2)^{2}(\mathrm{d}u^2 + \mathrm{d}v^2)$. The first fundamental form admits a one-parameter family of isometries $\psi_{t} : S \to S$ defined by $\psi_{t}\left(u, v\right) = \left( u\cos t - v\sin t , u\sin t + v \cos t\right)$, but Enneper's surface is not a surface of revolution. – THW Oct 25 '18 at 00:56
  • Off hand I do not know the conditions on when a first fundamental form with a one parameter family of isometries can be realized as a surface of revolution in three-dimensional space. I guess that this would amount to an intrinsic isometry also being realized extrinsically in an embedding. I will give it some thought if I get a moment, but I suspect that the answer must be known. – THW Oct 25 '18 at 00:56