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In Example 2.39. in Hatcher's Algebraic Topology (for the $3$-dimensional torus) he states:

...the local degree is $+1$ at one of these points and $−1$ at the other, since the restrictions of $\Delta_{\alpha\beta}$ to these two faces differ by a reflection of the boundary of the cube across the plane midway between them, and a reflection has degree $−1$.

I'm not really sure how to see the degree is $+1$ in the front face and $-1$ in the back one. What is the boundary of the cube? I thought this is defined for $\Delta$-complexes, but here it's not a $\Delta$-complex.

How am I suppose to understand the above statement in terms of the 'boundary' of the cube (whatever that means here).

Edit: I'm beginning to think that there's an implicit diagonal on each face, as appears in the $2$-dimensional torus at the very beginning of $\Delta$-complexes (p. 102) which enables us to discuss boundaries as we defined them for $\Delta$-complexes. I think, then, (denoting lower triangle on the front as $L$ and the upper as $U$) that on the front, the boundary is $L-U$, while on the back we have $-L+U$ since the orientation on the back within each of the two triangles in counter to what's on the front. Is that correct?

(Related posts that do not answer this question are here and here)

Anon
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  • This is very hand wavy, what a shame. It would be nice if he could make more of his computations explicit – FShrike Feb 13 '23 at 13:43
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    @FShrike Couldn't agree more. – Anon Feb 13 '23 at 13:49
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    This is based on Prop 2.30 and the diagram at the top of p. 136. The attaching map from $S^2$ to the 2-skeleton is locally a homeomorphism, and so you have to compute $H_2(U_i, U_i-x_i) \to H_2(V, V-y)$ where $y$ is, say, a point in the center of a face of the cube and $x_1$ and $x_2$ are its preimages in $S^2$. The choice of a generator of $H_2(U_i, U_i-x_i)$ corresponds to an "orientation" on $S^2$ — choose a little circle around each $x_i$, going counterclockwise (so it's consistent at each point) when viewed from the outside. Then what happens to that orientation around $y$? – John Palmieri Feb 13 '23 at 18:43
  • The cube is a $3$-disk, with its boundary a $2$-sphere. Whether the front face has degree $1$ or $-1$ depends on choices, but the two local degrees must have opposite signs because the local maps differ (compositionally) by a reflection of this boundary $S^2$ (that switches the front and back faces of the cube in a way that is consistent with the identifications). – ronno Feb 28 '23 at 16:42

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