Question

Differential Geometry:

1) Give an example of two vector fields X,Y E X(R3) such that for almost all p ER3 the three tangent vectors Xp, Yp, [X,Y]

Answer 1

Part 1)

Consider the standard coordinates
z'fr)
on $\mathbb{R}^3$ . Correspondingly we have the standard vector fields,
2x,y,
giving a framing for the tangent space $T\mathbb{R}^3$ . Denote the vector fields,
X = a,
and
Y = 0, +ra.
. We check,

[X,Y] = [8., 2y + x? a.] = [@z, Oy] + [2., 22 .] = 0+ 2.0. = 2:00:

Now observe that for any
PER
, where the coordinate is non-zero, we have that
X , Yp, (X,Y)
is linearly independent. Whereas, for points
p= (0, P2, P3)
we have,
[X,Y), = 2.0.0.1 = 0
. Thus,
X , Yp, (X,Y)
does not span the tangen space $T_p\mathbb{R}^3$ for
p= (0, P2, P3)
. Clearly such points are just the -plane, which is a measure zero set. Hence for almost all points we have that
X , Yp, (X,Y)
is indeed a basis.

Next we compute,
X, X,Y)= (0., 2..= 20
. Then clearly for points
p= (0, P2, P3)
, we have that
[X,[X,Y]]= 20:1p
. Hence,
X,Y, X, X, Y)
forms a basis for such points.

Note: The vector fields in this example spans the martinet distribution, given as the kernel of the one form
"Cz2 - zp = 0
on $\mathbb{R}^3$ .

Part 2)

Suppose we have any such vector fields as in part 1. If possible, be tangent to some surface
SCR3
. Thus we have,
X, Y, ETS
for
pes
. But we also have,
X,YET,
and
[x, (X,Y]], E TAS
. Hence, we see that
X , Yp, (X,Y)
or
X,Y, X, X, Y)
can never be a basis of of $T_p\mathbb{R}^3$ for
pes
. This contradicts the assumption of part 1.

Hence no such pair of vector fields (as in part 1) can be tangent to any surface

SCR3

Hope this helps. Feel free to comment for any further clarifications. Cheers!