Why is $\alpha(t)=(t^3,t^2)$ not an immersion?

Question

$$\alpha: \mathbb{R}\to\mathbb{R}^2\quad \alpha(t)=(t^3,t^2)$$ The condition for an immersion is that $\alpha'(t)=(3t^2,2t)$ is injective for all $t\in\mathbb{R}$.

The book seems to imply that there is a problem at $t=0$ but I don't understand why. At $t=0$ the tangent is undefined because if you take the limit from both sides you get opposite results. However that does not make $\alpha'$ non-injective.

What am I missing?

Answer 1

The condition for immersion is not that the function $\alpha'$ is injective. It is that, for each $t$, the linear transformation $\alpha'(t):\mathbb R\to\mathbb R^2$, given by $$ (3t^2,2t)\,s=(3t^2s,2ts). $$ is injective. At $t=0$, the linear transformation is the zero map, which of course is not injective.

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Part of the problem here is an awful convention that creeps up from calculus, and pre calculus, which is to call a function $f$ as $f(x)$. So, in this case, where $\alpha'$ is a function but $\alpha'(t)$ is a function for each $t$, the poor notation wreaks havoc with the inexperienced student.

Answer 2

The map is a submersion if the differential is injective. However the differential is a linear map.

For instance at a point $t_0\in\mathbb{R}$ the differential at $t_0$ is $d\alpha_{t_0}(x)=(3t_0^3,2t_0)x$.

This linear map is injective except at $t_0=0$ which gives the null map.

Answer 3

Keep in mind that a map between manifolds $f:M\to N$ is called an immersion at the point $p\in M$ when the differential $f_{*p}:T_pM\to T_{f(p)}N$ is $1-1$, that is when $\ker(f_{*p})=\{0\}$. $f$ is called an immersion when it is an immersion on every point of the domain manifold $M$.

Let's see if we can apply this to your map $a:\mathbb{R}\to\mathbb{R^2}$ with $a(t)=(t^3,t^2)$. For a point $p$ of $\mathbb{R}$, it is $a_{*p}:T_p\mathbb{R}\to T_{(p^3,p^2)}\mathbb{R}^2$ with $\displaystyle{a_{*p}\big{(}\frac{d}{dt}\vert_p\big{)}=\frac{dt^3}{dt}\vert_p\frac{\partial}{\partial x}|_{(p^3,p^2)}+\frac{dt^2}{dt}\vert_{p}\frac{\partial}{\partial y}\vert_{(p^3,p^2)}=3p^2\frac{\partial}{\partial x}\vert_{(p^3,p^2)}+2p\frac{\partial}{\partial y}\vert_{(p^3,p^2)}}$,

so for any vector $\lambda\frac{d}{dt}\vert_p\in T_p\mathbb{R}$:

$\displaystyle{a_{*p}\big{(}\lambda\frac{d}{dt}\vert_p\big{)}=3\lambda p^2\frac{\partial}{\partial x}\vert_{(p^3,p^2)}+2\lambda p\frac{\partial}{\partial y}\vert_{(p^3,p^2)}}$ Therefore, at the point $p=0$ you have $a_{*0}(v)=0$ for any $v\in T_0\mathbb{R}$. Hence $\ker(a_{*0})=T_0\mathbb{R}$, and $a$ is not an immersion at point $0$. Now if you exclude 0 and consider the domain of $a$ to be the submanifold $(0,+\infty)$, $a$ is indeed an immersion.