# Tangent bundle of a surface is a manifold

My differential geometry textbook defined the tangent bundle of a surface as the set of all tangent vectors to M at all points of M.

The abstract patches are also given :

$$y(p_{1},p_{2},p_{3},p_{4})=p_{3}x_{u}(p_{1},p_{2})+p_{4}x_{v}(p_{1},p_{2})$$

where $x$ is patch in surface $M$.

$(p_{3},p_{4})$ is vector part and $(p_{1},p_{2})$ is a point of application.

I checked that $y$ is one-to-one.

I also checked that $T(M)$ satisfies the covering property and the Hausdorff property.

Thus if I show that it satisfies the smooth overlap property, $T(M)$ will become manifold.

The smooth overlap property : For any patches $x,y$ in collection of patches $P$, the composite functions $y^{-1}x$ and $x^{-1}y$ are Euclidean differentiable.

I assumed that $x,z$ are patches in M and $y_{1},y_{2}$ are patches in $T(M)$

$$y_{1}^{-1}y_{2}(p_{1},p_{2},p_{3},p_{4})=y_{1}^{-1}[p_{3}x_{u}(p_{1},p_{2})+p_{4}x_{v}(p_{1},p_{2})]=(q_{1},q_{2},q_{3},q_{4})$$

Then we get

$$y(q_{1},q_{2},q_{3},q_{4})=q_{3}z_{u}(q_{1},q_{2})+q_{4}z_{v}(q_{1},q_{2})=p_{3}x_{u}(p_{1},p_{2})+p_{4}x_{v}(p_{1},p_{2})$$

Thus $q_{3}=p_{3}$, $q_{4}=p_{4}$, $z_{u}(q_{1},q_{2})=x_{u}(p_{1},p_{2})$, $z_{v}(q_{1},q_{2})=x_{v}(p_{1},p_{2})$.

In conclusion,

$$y_{1}^{-1}y_{2}(p_{1},p_{2},p_{3},p_{4})=((z_{u}^{-1}x_{u})(p_{1},p_{2}),p_{3},p_{4})$$ $$=(P_{1}((z_{u}^{-1}x_{u})(p_{1},p_{2}),P_{2}(z_{u}^{-1}x_{u})(p_{1},p_{2}),p_{3},p_{4})$$

where $P_{1},P_{2}$ is a projection function.

Since $P(z_{u}^{-1}x_{u})(p_{1},p_{2})$ are differentiable, $y_{1}^{-1}y_{2}$ is differentiable.

I think I solved it.

-