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I am beginning to learn algebraic topology. For a 2-simplex $\{v_0v_1v_2\}$, the boundary is the chain 1-simplexes: $\partial(v_0v_1v_2)=v_0v_1+v_1v_2+v_2v_0$, which naively and intuitively, I see as the triangle that bounds the 2-simplex, that triangle being made up of the edges (1-simplexes) of the little triangular surface element. So far, so good. Then, $\partial\partial(v_0v_1v_2)=v_0-v_1+v_1-v_2+v_2-v_0=0$, which is algebraically convincing, and means that "the boundary of the boundary is empty".

Is this intuitively (and naively) because the boundary has to "loop around" the initial simplex, that characteristic of being a "loop" necessarily implying that following the boundary gets you to where you started from, which is as if you had not moved at all? Or is there a better way to intuitively grasp what is going on with that $\partial\partial = 0$?

Frank
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  • @JWL Are they really related, though, except semantically? Because, for the boundary of a manifold, you have induced charts $\phi : U \cap \partial M \to \partial H^n = R^{n−1}$, on the boundary, so by definition, it is a manifold without boundary. Whereas, the boundary operator in homology orients the boundary in such a way that the orientations on the boundary of the boundary "cancel out." These ideas seem to be different. – Matematleta Jul 07 '19 at 03:23
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    @Matematleta I think you are right. Some things got messed up in my head. Indeed, the OP's question is about orientability. – JWL Jul 07 '19 at 03:28
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    The boundary of the triangle is topologically the same as a circle; a circle has no ends (unlike a line segment) so has empty boundary.... – Angina Seng Jul 07 '19 at 04:56
  • The boundary of a $2$-simplex is not a $1$-simplex, but a chain (=formal sum) of $1$-simplices. – Paul Frost Jul 07 '19 at 08:30
  • @PaulFrost - corrected my post. Thanks. – Frank Jul 07 '19 at 15:08

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