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I have a closed path in a smooth compact Manifold $C \subset M$, that consists of straight pieces, so $C = \alpha \circ \beta \circ ... \circ \delta$, where $\circ$ is the concatenation, and $\alpha = tx + (1-t)y$ , $\beta = ty+(1-t)z ,...$ for some $x,y,z,... \in M$ and $t \in [0,1]$. How would I change this path, to make it differentiable, assuming i dont have any self-intersections? I can choose a small neighbourhood of each of these $x,y,z,...$ where i can do whatever i want, to make it differentiable, and not change the homotopy. but how exactly do i change it, to make it differentiable?

PlatinTato
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In my answer I'm going to use $*$ as the concatenation operator, since I must use $\circ$ in its usual role as the composition operator.

You can reparameterize each of the paths $\alpha,\beta,\ldots,\delta : [0,1] \to C$ by precomposing each with a homeomorphism $f : [0,1] \to [0,1]$ whose derivative $f'(t)$ is continuous and such that $f'(0)=f'(1)=0$. The resulting composed paths $\alpha \circ f,$ $\beta \circ f,$ ..., $\delta \circ f$ are homotopic rel endpoints to $\alpha,\beta,\ldots,\delta$ respectively, because $f$ is homotopic to the identity rel endpoints. The resulting concatenation $(\alpha\circ f) * (\beta\circ f) * \cdots * (\delta \circ f)$ is continuously differentiable, and it is homotopic rel endpoints to the original path $\alpha * \beta * \cdots * \delta$.

Straightness of the pieces $\alpha,\beta,\ldots,\delta$ is not necessary for this argument, and might be impossible to achieve for a general smooth compact manifold $C \subset M$. Fortunately, this exact same argument works as long as each of $\alpha,\beta,\ldots,\delta$ is continuously differentiable.

Also, the assumption of no self-intersections is not necessary for this argument, it works regardless of whether $\alpha \circ \beta \circ \cdots \circ \delta$ has self-intersections.

Lee Mosher
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  • First of, thanks for the answer. i got a couple of questions though. First, could i choose $f = id_{[0,1]}$, therefor showing that the original curve $C$ is differentiable? this seems somewhat counterintuitive at best. Secondly, The image if this new path still is the same submanifold, therefore if the original curve C didnt have any selfintersections, the "new" same curve wouldnt have any either right? – PlatinTato May 10 '17 at 14:18
  • $id_{[0,1]}$ doesnt have derivate 0 on its boundary. I think i asked the question wrong. i was interested in $C$ beeing a differentiable submanifold $C\subset M$. thank you anyways. Thank you anyways for your answer – PlatinTato May 10 '17 at 14:23
  • By the way, I did intend that my answer applies under the assumption that $C$ is a differentiable manifold. It's a good question, which does come up sometimes. I think of your question like this (and this is the spirit of my answer): The concatenation operator need not result in a smooth path at a concatenation point, so how do you fix that? My answer is: reparameterize each piece to slow it down at its endpoints. – Lee Mosher May 10 '17 at 14:26
  • I see. I really do think i missstated the question. i am trying to prove, that if $C$ is piecewise straight, and has countably many "vertices", then i can make it a differentiable submanifold. – PlatinTato May 10 '17 at 14:30
  • Hah, yes, that is a different question. And much harder! – Lee Mosher May 10 '17 at 14:49
  • https://math.stackexchange.com/questions/2274855/making-a-piecewise-straight-curve-a-differentiable-submanifold in case you are interested in the new question. Might it be possible with bezier curves? i dont know :( – PlatinTato May 10 '17 at 14:51