I am having problems with a problem of my smooth manifolds class about submanifolds and derivation.
Let $f:\Bbb{R} \hookrightarrow \Bbb{R}$ a smooth map such that $\forall t \in \Bbb{R}:f(t) >0$ . Consider the set $M=\{(x,y,z,t) \in \Bbb{R}^4 : x^2 + y^2 + z^2 =f^2(t) \}$ .
- Prove that $M$ is a smooth submanifold of $\Bbb{R}^4$
- Find $a\in \Bbb{R}$ such that for all $p=(x,y,z,t) \in M$ it has $X_p \in T_pM$ where $$X_p=(xf'(t)+yf(t)) \cdot \frac{\partial}{\partial x}\Big|_p + (yf'(t)-xf(t)) \cdot \frac{\partial}{\partial y}\Big|_p + zf'(t) \cdot \frac{\partial}{\partial z}\Big|_p + af(t) \cdot \frac{\partial}{\partial t}\Big|_p$$
For 1, let $F: \Bbb{R}^4 \hookrightarrow \Bbb{R}, (x,y,z,t) \mapsto x^2+y^2+z^2 - f^2(t)$ .
F is clearly smooth and $\nabla _pF =(2x,2y,2z,-2f(t)f'(t))$, so $\nabla_p F = \overrightarrow{0} \Rightarrow (x,y,z)=(0,0,0)$ . If $p=(x,y,z,t) \in M=F^{-1}(0)$ then $x^2+y^2+z^2-f^2(t)=0$ and $f(t) >0$ because of how $f$ is defined. Hence, $x^2+y^2+z^2 >0$, so $(x,y,z) \neq (0,0,0)$, so $0$ is a regular value of $F$. Therefore, by the Regular Value Theorem $M=F^{-1}(0)$ is a smooth submanifold of $\Bbb{R}^4$ .
I wrote this proof that I think it is fine, but I would appreciate confirmation.
For 2, I am not sure if I am misunderstanding something but I think that $X_p$ does not depends of $a$ to be in $T_p M$ since is a linear combination of their base vectors, so any advice or correction would be great.
Thanks in advance :)
You are correct about the first part. Moreover, the regular value theorem tells you that $M$ is of codimension $1$, or in your case, of dimension $3$. In particular the tangent space has dimension $3$, and it is not true that all of $\partial_x,\partial_y,\partial_z,\partial_t$ are a basis of the tangent space at any $p\in M$. In fact, the span the (larger) tangent space to $\mathbb{R}^4$, of which $M$ is a codimension $1$ submanifold.
To answer your second question. Your manifold is defined by the relation $$F(x,y,z,t):=x^2+y^2+z^2-f(t)^2=0.$$ If you consider curves inside $M$ through some point $(x,y,z,t)$ and differentiate, you will find that the tangent vector $v$ to such a curve must satisfy (chain rule!): $$\nabla_p F\cdot v=0,$$ with $\nabla_p F$ the gradient that you -correctly- compute in your question. In the case of $F$, this boils down to $$(2x,2y,2z,-2f(t)f'(t))\cdot (v_x,v_y,v_z,v_t)=0.$$ Note that we are writing $v=v_x\partial_x+\dots+v_t\partial_t$ but, as we said above, not every choice of $v$ will satisfy the relation above. If you pick $X_p=(xf'(y)+yf(t),yf'(t)-xf(t),zf'(t),af(t))$ you will find that $$\frac{1}{2}(\nabla F\cdot X_p)=(x^2+y^2+z^2)f'(t)- af^2(t)f'(t).$$ Now $p\in M$ and this allows you to simplify the expression above ... Can you see for which $a$ it will be zero?