How Do You Determine Basis Vectors for the Tangent Space of a Surface in R^4?

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SUMMARY

The discussion centers on determining basis vectors for the tangent space of a surface parameterized in R^4 by the function \(\alpha(u) = (f_1(u), f_2(u), f_3(u), f_4(u))\). The user initially confuses the dimensionality of the tangent space, which requires two vectors for a surface, not three. The correct approach involves calculating the partial derivatives of the parameterization, leading to three linearly independent vectors that form a basis for the tangent space. The rank of the Jacobian matrix confirms the independence of these vectors.

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  • Understanding of surface parameterization in R^4
  • Knowledge of partial derivatives and their application
  • Familiarity with the Jacobian matrix and its properties
  • Concept of tangent spaces in differential geometry
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Alteran
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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:
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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.
 
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:
I believe it makes most sense to take {da/du1(p), ..., da/du3(p)}
 
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?
 
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.
 
Yes, rank of Jacobian matrix is equal to 3, so these vectors are linearly independent. Looks like they are basis vectors.

Thanks for support.
 

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