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I read that a curve $\gamma:S^1 \to \mathbb{R}^2$ is an immersion iff it's regular. So it's an immersion iff $\gamma'(t) \neq 0$ for all $t \in S^1$.

An immersed curve is one whose derivative is an injective linear map. When the domain is single-variabled, this is the same as requring $\alpha'(t) \neq 0$ for all $t$.

Is this right? But what about $t=0?$ Shouldn't $0$ make the derivative equal $0$? Why isn't this included in this characterisation (since regular means non-zero everywhere)?

How does one check that such a curve is immersive? If $\gamma(t) = (e^t, 1)$, which I know is regular, how do I check that the derivative is an injective linear map? Clearly the "derivative" in this context isn't referring to $\gamma'(t)$ since that is not linear.


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Why should $0$ make the derivative equal $0$? – Ben Millwood Jul 5 '12 at 11:50
up vote 4 down vote accepted
  1. In one-variable calculus one introduces the derivative of a function $f:{\mathbb R}\to{\mathbb R}$ as derivative at a point $t_0$ by defining $$f'(t_0):=\lim_{t\to t_0}{f(t)-f(t_0)\over t-t_0}\ .\qquad(1)$$ This is a number. In a second step one envisages this process performed at all points $t_0\in{\rm dom}(f)$, which leads to the idea of the derivative being a function $$f':\quad {\mathbb R}\to{\mathbb R},\qquad t_0\mapsto f'(t_0)\ .$$ This works as well for vector-valued functions of a single variable $t$, e.g. for the function $\gamma:\ t\mapsto(e^t,1)$. Here one gets $\gamma'(t)=(e^t,0)$.

  2. When one has more than one independent variables the derivative $d{\bf f}({\bf p})$ of a function $${\bf f}:\quad {\mathbb R}^n\to{\mathbb R}^m$$ at a point ${\bf p}$ can no longer be defined as a limit in the form $(1)$. Instead $d{\bf f}({\bf p})$ is a linear map mapping the tangent space at ${\bf p}$ into the tangent space at ${\bf q}={\bf f}({\bf p})$. The matrix of $d{\bf f}({\bf p})$ is nothing else but the Jacobian matrix of ${\bf f}$ at ${\bf p}$. The map ${\bf f}$ is an immersion if this matrix has rank $n$ at all points ${\bf p}$ in the domain of ${\bf f}$.

  3. What we have said in 2. also makes sense in the one-variable case, i.e. for $n=1$: $$t\mapsto{\bf f}(t)=(f_1(t),\ldots,f_m(t))\ .$$ In this case the Jacobian at a point $p$ in the domain of ${\bf f}$ has just one column given by $[f_1'(p),\ldots,f_m'(p)]^T$, and it has rank $1$ iff ${\bf f}'(p)\ne{\bf 0}$, where ${\bf f}'(p)$ is the derivative of ${\bf f}$ at $p$ considered in 1. In your example $\gamma'(t)\ne(0,0)$ for all $t\in{\mathbb R}$, so $\gamma$ defines an immersion.

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Thank you, great answer. If I see a curve, is there a way to visually tell that it's an immersion? – soup Jul 5 '12 at 14:37
@soup: The map $\gamma(\cdot)$ producing a curve is an immersion or not. A curve with "corners" cannot be an immersion. The map $t\mapsto (t^3,0)$ is not an immersion, but you cannot see this from looking at the curve. – Christian Blatter Jul 5 '12 at 14:59

The basic problem here is a confusion of terminology around "derivative".

The basic idea of differentiation in one dimension is finding a tangent to a graph at the point. A tangent can be viewed as a linear approximation of the graph, in the following (not necessarily obvious) sense: $$f(x + h) - f(x) \approx hf^\prime(x)$$ The function that you are really linearly approximating is not $f$ as such but the difference between $f(x)$ and $f$ near $x$. Since this will be zero when $h$ is zero, it seems much more plausible that we could approximate it in this way (rather than $f$ which can be whatever it likes).

In higher dimensions, the derivative of $f$ at $x$ is defined to be a linear map $A$ such that $$f(x + h) - f(x) \approx A(h)$$ And in fact, the 1D case fits here, since $h\mapsto ah$ is linear for any constant $a$. The problem is that in fact every linear map in 1D is of the form $h\mapsto ah$, so there is a direct correspondence, in the one-dimensional case, between scalar constants and linear maps. So there ends up being confusion in our definitions whether the derivative of $f$ at $x$ is a number or a function, and I think it is that confusion that you are coming up against.

In this context, don't think of $f^\prime$ as the derivative of $f$, think of it as a function that takes an argument $x$ and gives you the $a$ such that the derivative of $f$ at $x$ is "multiply by $a$". This is a linear map, which is injective exactly when $a\not=0$. The key is to understand that $f^\prime$ isn't a linear map itself, it produces linear maps for each $x$.

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Thanks a lot for the answer. – soup Jul 5 '12 at 14:38

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