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    Edge identifications of $\Delta^3$ producing Klein bottle, torus, $S^2$, $\mathbb{RP}^2$

    E-HAT-2.1-2

    Area
    Topology
    Topics
    Homology, Simplicial Complexes, Surfaces
    Status
    draft

    Show that the \(\Delta\)-complex obtained from \(\Delta^3\) by performing the edge identifications \([\nu_0, \nu_1] \sim [\nu_1, \nu_3]\) and \([\nu_0, \nu_2] \sim [\nu_2, \nu_3]\) deformation retracts onto a Klein bottle. Find other pairs of identifications of edges that produce \(\Delta\)-complexes deformation retracting onto a torus, a 2 sphere, and \(\mathbb{RP}^2\).

    Solution

    Let the vertices of the tetrahedron be \[ \nu_0,\nu_1,\nu_2,\nu_3. \] The union \[ Q=[\nu_0,\nu_1,\nu_3]\cup[\nu_0,\nu_2,\nu_3] \] is a square split by the diagonal \([\nu_0,\nu_3]\), with boundary traversed as \[ \nu_0\to\nu_1\to\nu_3\to\nu_2\to\nu_0. \]

    1. The tetrahedron deformation retracts onto \(Q\) in a way compatible with identifications made only among the four boundary edges of \(Q\).

    The two omitted faces \[ [\nu_0,\nu_1,\nu_2], \qquad [\nu_1,\nu_2,\nu_3] \] form an open book on the edge \([\nu_1,\nu_2]\) opposite the diagonal \([\nu_0,\nu_3]\). Collapse each line segment transverse to \(Q\) linearly onto \(Q\). This fixes \(Q\) pointwise and gives a deformation retraction. Since all edge identifications below involve only edges of \(Q\), the retraction descends to the corresponding quotients.

    2. The identifications in the problem,\[ [\nu_0,\nu_1]\sim[\nu_1,\nu_3], \qquad [\nu_0,\nu_2]\sim[\nu_2,\nu_3], \]produce the Klein bottle.

    Call the first edge class \(a\) and the second \(b\). Around the square boundary \[ \nu_0\to\nu_1\to\nu_3\to\nu_2\to\nu_0 \] the four directed sides become \[ a,\ a,\ b^{-1},\ b^{-1}. \] Thus the quotient polygon has boundary word \[ a^2b^{-2}. \] After replacing \(b\) by \(b^{-1}\) this is the standard nonorientable genus-two word \[ a^2b^2, \] which is a Klein-bottle polygon. By 1 the full tetrahedral quotient deformation retracts onto this surface.

    3. A torus is obtained by the pairings\[ [\nu_0,\nu_2]\sim[\nu_1,\nu_3], \qquad [\nu_0,\nu_1]\sim[\nu_2,\nu_3], \]with the displayed orderings preserved.

    Let the first edge class be \(b\) and the second \(a\). The square boundary word is then \[ a\,b\,a^{-1}\,b^{-1}, \] the standard torus polygon.

    4. A \(2\)-sphere is obtained by the pairings\[ [\nu_0,\nu_2]\sim[\nu_3,\nu_2], \qquad [\nu_0,\nu_1]\sim[\nu_3,\nu_1]. \]

    With suitable labels the square boundary word is \[ a\,a^{-1}\,b\,b^{-1}. \] The two adjacent inverse pairs cancel by folding the corresponding bigons, leaving a \(2\)-sphere. Equivalently, this is the usual two-disk decomposition of \(S^2\) after the cancellations.

    5. A projective plane is obtained by the pairings\[ [\nu_0,\nu_2]\sim[\nu_3,\nu_1], \qquad [\nu_0,\nu_1]\sim[\nu_3,\nu_2]. \]

    The quotient has two vertex classes, two edge classes, and one \(2\)-cell. One edge is a maximal-tree edge joining the two vertices. Collapse it. The remaining \(1\)-cell becomes a loop, and tracing the square boundary shows that the \(2\)-cell attaches by degree two around this loop. Thus the quotient has the standard CW structure \[ e^0\cup_{2}e^1\cup e^2 \] of \(\mathbb{RP}^2\).

    6. Hence the requested examples can be chosen exactly as in <1>2–<1>5.

    Each follows from the square retraction in 1 and the corresponding polygon identification.