# Second fundamental form of surface with diagonal metric

1. Jun 12, 2012

### Sina

Hello everyone,

Let $$r(u_i)$$ be a surface with i=1,2. Suppose that its first fundamental form is given as

$$ds^2 = a^2(du_1)^2 + b^2(du_2)^2$$

which means that if $$r_1 = ∂r/∂u_1$$ and $$r_2= ∂r/∂u_2$$ are the tangent vectors they satisfy

$$r_1.r_2 = 0$$

$$r_1.r_1 = a^2$$

$$r_2.r_2 = b^2$$

My aim is to is to calculate the second fundamental form (whose answer I know but is a boring expression so I won't write here). The most direct approach is to calculate the levi-civita connection coefficients from the metric and then calculate curvature and then scalar curvature which will be Gauss curvature upto a factor (since it is a surface). But this is very oafy calculation.

The other way is to somehow calculate the second fundamental form whose coefficients are given by $$(r_{ij},N)$$ where $$r_{ij} = ∂^2r/∂u_i∂u_j$$ and N is normal to the surface (generically given by) $$r_1 \times r_2$$ . Then using this and the metric one can calculate Gauss curvature (as the eigenvalues of G-1B where G is the first and B is the second fundamental form).

As a motivation I tried a particular simple example where the normal vector is r/|r| itself (for instance sphere). Then taking derivatives of the equation 0 = r1.r and using equalities (that come from the assumption of the metric) $$a.a_i = r_1.r_{1i}, b.b_i = r_2.r_{2i}$$ it is a matter of seconds to find the second fundamental form (in fact off diagonal terms vanish etc). However normal vector is not necessarily r itself in the case of general diagonal metric. All we can say is $$r_1.r_2 = 0$$ which again by derivation yields some equations but does not help me because I don't know any other way to express the normal vector other than r_1 x r_2 which is useless to do component calculations (if there was a way to turn products of the form $$(r_1 \times r_2).r_{ij}$$ into serioes of dot products that would also be enough but as far as I know there is no such formula and you can only exchange the places of factors). So my question is can you think of anyway to express N in some particular simple way in such a metric. I am really hoping that there should be a way to do this simply.

Thanks

Last edited: Jun 12, 2012
2. Jun 12, 2012

### Muphrid

I'm not sure why it is you think $r_1 \times r_2$ is useless for component calculations? You get a vector that can be expressed in terms of a basis.

3. Jun 12, 2012

### Sina

You mean

$$r_1 \times r_2 = ε_{ij}^{k} r_{1}^ir_{2}^je_k$$

where e is the basis for R3. Well without knowing what e^k is, I can not calculate the dot product of this with other vectors?

The reason why I gave the motivating example is in that case normal vector N is directly related to r and I can dig out its dot products with r_ij through the relations like

0 = ∂i(rj.r) and some other equations I gave above (where since r is orthonormal to tangent vectors we use (rj.r) =0 always).

4. Jun 20, 2012

### Sina

Hello Sina,

You can indeed doing it without resorting to theorema egregium.

Let r(u,v) be the surface with line element ds2 = a2du2 + b2dv2 and tangent vectors ru, rv. Then we have the equalities

ru.ru = a2
rv.rv = b2
ru.rv =0

Then we have the following

ruu.ru = aua
ruv.ru = ava
rvv.rv = bvb
ruv.rv = bub

Now we have the orthogonal basis

eu = ru/a
ev = rv/b
N = euxev (which we fortunately dont really need)

Then we have the second fundamental form coefficients

buu = ruu.N
buv = ruv.N
bvv = rvv.N

We can write each vectors in the orthogonal basis, for instance

ruu = (ruu.N)N + (ruu.ru)ru/a2 + (ruu.rv)ru/b2

and similiarly for the others. Using the equalities above we get

buuN = (ruu,N) = ruu + ava rv/b2 - aua ru/a2

bvvN = (rvv,N) = rvv - bvb rv/b2 + bub ru/a2

buvN = (ruv,N) = ruv + ava ru/a2 - bub rv/b2

Then from here we want to calculate the determinant of the second fundamental form:

buu.bvv - buv2.

But

buu.bvv = [(ruu.N)N].[(rvv.N)N] which by above is

ruu.rvv + avbv a/b + aubub/a

and buv2 = (ruv.ruv) - bu2 - av2

All it remains to compute are quantities like ruu.rvv and ruv.ruv ( infact when you expand the first one it cancels the second one in calculating the determinant). These can be easily calculated by taking further derivatives of the first relations (you will have to get derivatives until you see expressions like ruvv etc so be generous). Then

we have K = det(B)/det(G) where det(G) = b2a2 and the final result
is

K = -(1/ab)*((av/b)v + (bu/b)u)

5. Jun 23, 2012

### lavinia

I have never tried this but the metric gives you a covariant derivative. The difference

$\nabla_{X}$Y - X.Y is normal to the surface.

The calculation of the covariant derivative may be easier if the coordinates are orthogonal.

What about if the coordinates are geodesic normal coordinates?

6. Jun 23, 2012

### Sina

Hmm I guess the formula would be simpler because because metric is identity and symbols vanish. But I guess that would not then cover the most general case but only when a=b=1.

Your first idea might give another easy answer maybe but my whole aim was to save my self from calculating the symbols :) After calculating the symbols I guess it is not much of a work to find the curvature since the curvature tensor has few components. I believe it would be something like R1t12gt2 or something similiar.

The solution I have given to myself (forever alone :p) also seems quite easy to me since you dont really need to compute N.

edit: the last thing I wrote here (and deleted) only proves theorem egregium not this sorry

Last edited: Jun 23, 2012
7. Jun 23, 2012

### lavinia

I often solve for curvature this way.

Normalize the orthogonal coordinate vectors to be of unit length and compute the dual 1 forms, $\omega_{1}$ and $\omega_{2}$

Solve for the connection 1 form,$\varphi$, from the equations

d$\omega_{1}$ = $\varphi$^$\omega_{2}$

and d$\omega_{2}$ = - $\varphi$^$\omega_{1}$

Then d$\varphi$ = -KdS

In geodesic normal coordinates the calculation is easy.

8. Jun 23, 2012

### Sina

Hmm have never seen this way. I usually encountered connection forms for connections on general vector bundles never tried to use them for tangent bundle but it will be good food for taught thanks

Last edited: Jun 23, 2012
9. Jun 23, 2012

### lavinia

- In terms of connections on vector bundles, the connection 1 forms are the components with respect to a frame of a general connection on the dual tangent bundle to the surface. Since the connection is Levi-Civita, the 2x2 matrix of 1 forms boils down to a single non-zero form.

That is why one can speak of the connection 1 form.

- For a surface embedded in 3 space, the unit normal may be viewed as a map from the surface into the unit sphere.

The tangent bundle of the surface is the induced bundle of the tangent bundle to the sphere under this map.

The connection on the tangent bundle to the surface is the induced connection. So the curvature 2 form on the surface is the pull back of the curvature 2 form on the unit sphere which in turn, is just the volume element of the sphere. This is a way to see why the determinant of the Gauss map (or second fundamental form) is the curvature of the surface.

Last edited: Jun 23, 2012