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what is the meaning of a curve $C$ representing Identity in fundamental group?

Suppose $M$ is a closed 3-manifold, $C\subset M$ is a simple closed curve which represent identity in $\pi_{1}(M)$. Then $C$ bounds an immersed disk in $M$.

My question is:

When does it bound an imbedded disk in $M$?

I don't know about it at all. If you have any reference, please tell me. Thank you!

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Better for math.SE , no? The answer is a standard theorem. – Daniel Moskovich Nov 25 2011 at 12:58
No question that math.SE is better, but this does not look like homework, and it sounds like OP is trying to figure things out on his own... – Igor Rivin Nov 25 2011 at 14:06

Suppose that $K$ is a simple closed curve in $M^3$. I'll assume that $M$ is orientable, compact, and without boundary. Let $V$ be a closed regular neighborhood of $K$; so $V \cong S^1 \times D^2$ is a solid torus. Let $X$ be the closure of $M - V$; so $X$ is the exterior of $K$. Let $T = X \cap V$; so $T$ is a two-torus. So $\partial X = \partial V = T$ and $M = X \cup_T V$. Note that $T$ is a two-torus. Let $D \subset V$ be a meridian disk; that is, a disk of the form $\lbrace \mbox{pt} \rbrace \times D^2$.

As Igor indicates, the map $\pi_1(T) \to \pi_1(X)$ induced by inclusion has a kernel if and only if there is a embedded disk in $E \subset X$ with boundary on $T$. If $\partial D$ and $\partial E$ meet once then $K$ bounds a disk in $M$.

To recap: the knot $K$ bounds an embedded disk in $M$ if and only if

1. the map from $\pi_1(T) \to \pi_1(X)$ has kernel and
2. the curve that dies ($\partial E$) meets the meridian $(\partial D$) exactly once.
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 To Sam: So it is not clear whether $K$ bounds an embedded or not. – yanqing Nov 26 2011 at 1:17 @Yanqing - My answer gives a necessary and sufficient condition for $K$ to bound an embedded disk in $M$. I'll restate this at the end of the answer. – Sam Nead Nov 26 2011 at 12:01 To same: Thanks a lot! – yanqing Nov 27 2011 at 5:21
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Dehn's Lemma (= Papakyriakopoulos' Theorem) asserts: if C represents 0 in $\pi_1M$ and if C is a simple closed curve, then C bounds an embedded disk.

The assumption on C being a simple closed curve is obviously necessary: multiples of C will not bound embedded disks.

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I think you are assuming that $C$ lies on the boundary of $M$, while the question does not make this assumption. – Kevin Walker Nov 25 2011 at 18:38
To Kevin: I assume that $M$ is a closed 3-manifold. – yanqing Nov 26 2011 at 1:09

Sorry, I don't know how to use comment (perhaps I can't use it). This is my comment to Sam: where do you use $[K]=0$ in $\pi_1{M}$? I feel that your answer is an explain of Dehn's lemma. I guess that "$[K]=0$" visual meaning is "$K\subset D^3 \subset M$" where $D^3$ is a 3 ball(similar to knot in $S^3$).

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 Being interior to a ball is not the same as being trivial in the fundamental group. In fact, it is a theorem of Bing that M is the 3-sphere if and only if every knot is interior to a ball. So, the Poincare conjecture would follow from your "visual meaning." – Richard Kent Nov 27 2011 at 21:45 Also, you need more reputation to comment. This is to prevent you from commenting until you've been around a while. – Richard Kent Nov 27 2011 at 21:47 @binyu : Are you from tongji university, Shanghai? – yanqing Nov 28 2011 at 5:19 @Richard: I don't understand the reason you say "Being interior to a ball is not the same as being trivial in the fundamental group." By your following argument, I can't see this. (since Poincare conjecture is a theorem now. This doesn't give a contradiction). @yanqing: yes – Bin Yu Nov 28 2011 at 5:59 I am saying that if being interior to a ball were the same as being trivial in the fundamental group, then you would have a new proof of the Poincare conjecture, which suggests that the assumption is false. The fact that they are not the same is easy to see: one component of the Whitehead link is null homotopic in the exterior of the other, but is not interior to a ball there. – Richard Kent Nov 28 2011 at 14:31
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