Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading R.E.Gompf and A.I. Stipsicz, 4-Manifolds and Kirby Calculus.At pp.100-101,the authors say "isotopy classes of framings."I don't know how to determine isotopy to framings.Please tell me about this.

share|cite|improve this question

Let $$\varphi_0, \varphi_1: V^k \hookrightarrow M^n$$ be two embeddings of the $k$-manifold $V$ into the $n$-manifold $M$, both with trivial normal bundle. Furthermore, let $f_0$ be a framing of the normal bundle $\nu(\varphi_0: V \hookrightarrow M)$, and similarly let $f_1$ be a framing of the normal bundle $\nu(\varphi_1: V \hookrightarrow M)$. Then $(\varphi_0, f_0)$ and $(\varphi_1, f_1)$ are two framed embeddings of $V$ into $M$.

The usual definition of an isotopy between the embeddings $\varphi_0$ and $\varphi_1$ is a homotopy $$F: V \times [0,1] \longrightarrow M$$ between $\varphi_0$ and $\varphi_1$ such that $F(-,t)$ is an embedding for all fixed $t$.

Note that by the Tubular Neighborhood Theorem, a normal framing $f$ of an embedding $$\varphi: V^k \hookrightarrow M^n$$ with trivial normal bundle uniquely determines an embedding $$\tilde{\varphi}: V^k \times D^{n-k} \hookrightarrow M^n$$ up to isotopy, where $D^{n-k}$ is the $(n-k)$-dimensional disk. Then we say two framed embeddings $(\varphi_0,f_0)$ and $(\varphi_1,f_1)$ are isotopic if the induced embeddings $$\tilde{\varphi}_0, \tilde{\varphi}_1: V^k \times D^{n-k} \hookrightarrow M^n$$ are isotopic as embeddings.

share|cite|improve this answer
Any chance you know how to find the integer number associated to a framing? – user99680 Feb 2 '14 at 4:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.