# Stuck with basis for tangent space

1. May 2, 2006

### Alteran

Stuck with problem:

There is a local surface $$\alpha(u) = (f_1(u), f_2(u), f_3(u), f_4(u)) \in R^4$$. I need to find basis vectors of tangent space on that surface in some point p. It is not difficult to calculate tangent space for that surface, we should choose some curve on the surface and then it's derivative, but how to find 3 vectors that will be a basis for tangent space? Is it Frenet trihedron?

Can anyone give me a hint? Should be easy.

Last edited: May 2, 2006
2. May 2, 2006

### mathwonk

your question makes no sense. a surface cannot be poarametrized by one variable u, that would bea curve. and then of couirse a surfacxe will have 2 vectors in a basis for its tangent spoace not 3.

if you real get a parametrization, then just carry over a basis for the tangent space of the coordinate space to your surface via the derivative of the parameter map.

3. May 2, 2006

### Alteran

ok, may be I did not specified extra data:
so:
$$u=(u_1, u_2, u_3)$$
and specifically
$$f_1(u)=sinu_1 sinu_2 sinu_3$$
$$f_2(u)=cosu_1 sinu_2 sinu_3$$
$$f_3(u)=cosu_2 sinu_3$$
$$f_4(u)=cosu_3$$

So basis of my tangent space will be a set of 3 vectors, tangent vector, normal vector, binormal vector? or I am wrong?

Last edited: May 2, 2006
4. May 2, 2006

### AKG

I believe it makes most sense to take {da/du1(p), ..., da/du3(p)}

5. May 2, 2006

### Alteran

Jacobian matrix is known, so I believe I can computate these derivatives by using that matrix, by summarize each row in columns (there are 3 columns denoted to each $$u_n$$ and 4 rows denoted to $$f_m(u)$$). By that procedure I will get 3 vectors, and they will be the basis of tangent space?

6. May 2, 2006

### AKG

Well, you don't need to know the Jacobian a priori, you can just the partial derivatives I mentioned (and in the process, you will end up pretty much computing the Jacobian). For example:

da/du1 = (df1/du1, ..., df4/du1) = (cosu1sinu2sinu3, -sinu1sinu2sinu3,0,0)

So if p = (p1, ..., p3), then:

da/du1(p) = (cosp1sinp2sinp3, -sinp1sinp2sinp3,0,0)

Do the same for the other partials, and you should get three vectors which will be a basis for the tangent space. How have you defined the tangent space? I would almost expect that, by definition, the tangent space is defined to be the span of these partial derivatives, i.e. it is the image of the Jacobian matrix. You can check that the vectors you get are linearly independent.

7. May 2, 2006

### Alteran

Yes, rank of Jacobian matrix is equal to 3, so these vectors are linearly independent. Looks like they are basis vectors.

Thanks for support.