What are...simplicial and singular homology?
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- čas přidán 27. 07. 2024
- Goal.
Explaining basic concepts of algebraic topology in an intuitive way.
This time.
What are...simplicial and singular homology? Or: Cycles modulo boundaries.
Nonsense.
The usual "of-by-one-error" around 4:04 - one should start counting at zero since a notation involving n symbols encodes something n-1 dimensional. I always get this confused, my bad.
A lot of nonsense this time.
My last slide is a good illustration why you want a computer to do the calculations. The homology calculation is just wrong, my bad. It should be 1) Z in n=0, Z/2Z if n odd, and 0 otherwise; 2) Q for n=0 and 0 otherwise; 3) Z/2Z always.
Not nonsense, but still beware.
In my head I am always using cohomology, and on my last slide I am using cohomological notation (C_0 on the left, delta maps pointing leftwards), while one should use the homological notation (C_0 on the right, delta maps pointing rightwards). That doesn't make a difference, that is convention anyway, but still be careful here.
Not nonsense, but still beware.
In step 3, "We get a chain complex:" is what is supposed to be written there.
Disclaimer.
Nobody is perfect, and I might have said something silly. If there is any doubt, then please check the references.
Disclaimer.
These videos are concerned with algebraic topology, and not general topology. (These two are not to be confused.) I assume that you know bits and pieces about general topology, but not too much, I hope.
Slides.
www.dtubbenhauer.com/youtube.html
Website with exercises.
www.dtubbenhauer.com/lecture-a...
Simplicial homology.
en.wikipedia.org/wiki/Simplic...
www.fields.utoronto.ca/progra...
Singular homology.
en.wikipedia.org/wiki/Singula...
Z versus other rings.
en.wikipedia.org/wiki/Univers...
Pictures used.
en.wikipedia.org/wiki/Simplic...
Hatcher’s book (I sometimes steal some pictures from there).
pi.math.cornell.edu/~hatcher/...
Always useful.
en.wikipedia.org/wiki/Counter...
Mathematica.
demonstrations.wolfram.com/Si...
#algebraictopology
#topology
#mathematics
loved how instead of presenting the formal definition first, you explained the intuitive way of looking at it. way clearer than how we approached the topic in class!
Thanks for the feedback!
Intuition is key in mathematics (and probably not just in math ;-)), and I certainly find it easier to first understand the “Why” before the “How”. That might not work for everyone, I am sure, so it is always good to get feedback!
A lecture is of course still somewhat different than a CZcams video. I kind of feel it should be CZcams < Lecture < Book when it comes to rigor, and vice versa for intuition. (Obviously a book still shouldn’t abandon intuition. But what I rather want to say is that a book has “more time” to be rigorous.) Its great to know that this approach works for someone else as well!
Thank you for making these videos. This one was very helpful for me since my prof glosses over so many details. This made so much more sense!!
Glad that the video was helpful! Homology looks very scary at the first sight (and is actually, at least in its general definition), so I hope that the video helped clarifying that the ideas are actually quite natural. Making them formally precise is a pain, but otherwise homology rocks!
Good job! I'm trying to read more into this to get a much deeper understanding of chebyshev filters. And of course we also have butterworth filters and others and I think that understanding these plynomials is the key to understanding the "why" and "how" of the working and of these filters that are commonly seen in signal processing.
Thanks. Might I ask how Chebyshev filters are related to homology. Never heard about that, sounds fun!
Thank you uploading these videos about algebraic topology. I've got an exam next week and they have been very useful so far 🙏
Awesome, I hope the videos will turn out to be helpful for the exams.
For my experience, taking and giving exams, I think the biggest influence in the end is “random noise”. In other words: Good luck!
Now I can get back to the abstract constructions with more ease.
Thanks!
Welcome!
I didn't get when "homology=kernel-rank" is applicable and when not.
So it does hold for all fields, but not for rings like Z?
That is correct: Over fields you can play the kernel-rank calculation (any field works - just remember to calculate "dimension" over the given field). Over general rings such as Z we do not have a good notion of "dimension", so the calculation does not work. Too bad.
Another way to think about it is: If kernel-rank would work over Z, then it would work over any field by specializing coefficients, and the result would be independent of the field. However, see my comment in brackets above, the calculation depends on the field. So it can't work over Z.
In summary, over fields its always kernel-rank, but over general rings more work is required.
Extremely good video. Easy and intuitive introduction. Question: Why are you writing Q^i? What‘s the idea behind that?
Not quite sure what you mean, and I hope this answers your question:
I need some ground field and I do not want to think about finite fields right now, so I take Q. Why? Well because it turns out that there is no essential difference between rational, real or complex numbers for homology, so I just take the easiest field. No other reason - you can easily work over R or C.
Moreover, I would like to have vector spaces with bases given by vertices, edges, faces and so on. The only thing I do now is that I choose a coordinate system such that I can identify vertices, edges, faces and so on with the standard vectors. In other words, I get Q^ci with c0=number of vertices, c1=number of edge, c2=number of faces and so on, and standard vectors correspond to vertices, edges, faces and so on.
And, thanks for the kind words ;-)
Is the statement 'working over the integers one can specialize to any field' related to how there is always a homomorphism from Z to any ring R? If so, what is known about how that homomorphism affects the homology groups when considered over R?
Good question:
- Yes, there is always a map Z->R, which implies that you can always extend coefficients from Z to R (via tensor products);
- For the chain complex that is reasonably easy, as on my last slide - just tensor with R instead of Q resp. Z/2Z (I tensored with Q resp. Z/2Z);
- For the homology this is somewhat tricky - you need to carefully analyze what happens to the boundary maps. This is captured by the universal coefficient theorem: en.wikipedia.org/wiki/Universal_coefficient_theorem
Great video! How do you find the 4x5 matrix and vector 6:47?
My intuition: For each vertex (rows), place a +1 if an edge (columns) has an arrow towards that vertex, place a -1 if the arrow points away from that vertex.
And for the vector (representing the face), place a +1 if an edge's arrow (row) goes in a counter-clockwise direction, and -1 if it's clockwise.
Excellent question, these matrices are key!
In short, your description is exactly right. Here is my version of saying the same:
- First, to even write a matrix we need to choose ordered bases. In other words, we order the simplices in some way (any order works) - I have chosen the order in the top right of the slide.
- The matrices are now given by taking any simplex and put a +-1 for every boundary component in its corresponding column.
- For example, the third column of the big matrix is the signed boundary of [0,3]. How can you see this? Well, note that my rows are ordered [0], [1], [2] and [3] and the boundary of [0,3] is [0] and [3] with [3] getting a sign.
- Note the silly typo in the top left diagram: all numbers should be shifted down by one so that I start counting at 0. The 2-simplex is then [0,1,2], and its boundary is [0,1], [1,2] and [2,3], with the latter read backwards so it gets a sign.
I hope that helps!
can you just tell me how it is that matrix maps the edges to faces? i cant see it at all
Sure, I give it a go! Sometimes it helps to be spelled out, so here we go:
Take the 5x4 matrix on slide 2. In the simplex above it you see five edges (they correspond to the columns of the matrix) and 4 vertices (they correspond to the rows of the matrix).
Now a matrix needs a choice of order on a basis. Here is my choice for that matrix: Basis of edges {[1,2],[1,3],[1,4],[2,3],[3,4]} and basis of vertices {[1],[2],[3],[4]}.
Cool. Now the matrix tells you for example that the second column is (1,0,-1,0) = 1*[1] + 0*[2] + (-1)*[3] + 0*[4] which is the image of the edge [1,3] under the matrix. The nonzero scalars are in front of the basis vectors for the vertices associated to [1,3]. (Note that the orientation determines the sign.)
I hope that helps!
is the simplicial complex a torus?
Maybe you got a bit confused? It is certainly my fault if you got confused, so let me clarify:
A simplicial complex is a certain type of structure on a space, if that structure exists. Every torus can be realized as a simplicial complex, but the converse is not true. (And that is good, since simplicial complexes should be a large class of spaces.) For example, a triangle is (can be realized as) a simplicial complex, but is certainly not a torus.
Sorry if that wasn't clear.