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I am reading a book on homological algebra. In order to determine the homology group of the $2$-dimensional projective space, the author identifies the space with the southern hemisphere of $S^2$, together with $s/\{ \pm 1 \}$, $s$ denoting the equator. $\mathbb{PR}^2 = \mathbb{R}^2 \cup \ell_{\infty}$.

That is:

$P \in \mathbb{R}^2$ is identified with $P' \in S^2$. $A$ and $B$ on $s$ represents the same point in $\mathbb{PR}^2$.

Then, this hemisphere is homeomorphic to a disk, and this disk is transformed continuously to a rectangular:

And this can be simplified to:

I can understand the first two parts, but I don't know how does the simplification take place.

Would you please give me some instructions? Many thanks?

These pictures are posted on a Chinese website, and I hope they can be seen to everyone.

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I don't know what simplified means, but your third picture does not give a triangulation of $P^2$ (because the two $2$-simplices intersect in a bad way), yet the second one does. –  Mariano Suárez-Alvarez Sep 10 '11 at 16:53
2  
The last picture is a Delta complex (or a CW complex). –  Grumpy Parsnip Sep 10 '11 at 17:10
    
Thank you @Mariano Suárez-Alvarez for your comment. I don't the simplification either. I just translated the contents from Chinese. –  ShinyaSakai Sep 11 '11 at 12:41
    
@Jim Conant: Thank you very much. Are Delta complex and CW complex the same? I found the definition for the latter, but had no idea of the first. Also, my textbook calls it a generalized simplicial complex. I am confused by the nomination... –  ShinyaSakai Sep 11 '11 at 12:46
    
Allen Hatcher defines delta complexes in his book "Algebraic Topology," which is available free from his webpage. A delta complex is like a simplicial complex, but the boundary of a simplex does not have to have distinct faces. –  Grumpy Parsnip Sep 11 '11 at 13:52

1 Answer 1

up vote 1 down vote accepted

All that is going on is a consolidation of adjacent triangles to give cells. As @Mariano and @Jim Conant pointed out in the comments, picture 2 is a triangulation, while picture 3 is an equivalent CW-decomposition.

Consider first that every point interior to the rectangle represents a distinct point of $\mathbb{R} P^2$ (this follows directly for the characterization given above in terms of the hemisphere). Now consider the points on the left side of the rectangle. In order from bottom to top, we have $P_0, P_1, P_2, P_3$. Taken together, we may label the segment $\overline{P_0P_3} =: a_1$. Indeed, the direction is important, so $a$ is labeled (in the third picture) together with an arrow to indicate the specified orientation of $a_1$. Notice that the right side of the rectangle in the second picture has the same vertices $P_0, P_1, P_2, P_3$, taken from top to bottom now. Again, $\overline{P_0P_3} =: a$, but the orientation is opposite. We must think of the left and right sides of the rectangle as containing the same points, identified is such a way as to make the arrows go the same direction. A similar analysis identifies the top and bottom by $\overline{P_3P_4P_5P_0} =: a_2$, but with opposite orientations. Finally, $a_3 := \overline{P_0P_8P_{10}P_7P_0}$ is included in the third picture to make sure the result is a "Delta complex." (very much like a triangulation, except that one may identify faces in more flexible ways). Note, it is a standard result (see Hatcher, e.g.) that the Delta-complex defines the same homology as the original triangulation of a space.

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Many thanks for your answer. Are CW-complex and delta complex the same? Would you please give me a reference on the result that the Delta-complex defines the same homology as the original triangulation of a space? I am sorry I searched on the web but didn't find much useful on Delta-complex. Thank you very much. –  ShinyaSakai Sep 11 '11 at 12:58
    
Hatcher has a very readable explanation, starting in section 2.1. The book is available at math.cornell.edu/~hatcher/AT/ATpage.html In a few words, every $\Delta$-complex has a subdivision which is simplicial. –  Shaun Ault Sep 19 '11 at 15:57
    
Also, $\Delta$-complexes can be regarded as CW-complexes, though not necessarily the other way around. –  Shaun Ault Sep 19 '11 at 15:58

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