# Questions tagged [homology]

Homology is a general way of associating a sequence of algebraic objects such as abelian groups or modules to other mathematical objects such as topological spaces.

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### Homology groups of moduli of parabolic bundles with fixed determinant

I am looking for the Homology groups of the moduli space of stable parabolic bundles over a smooth projective curve with fixed determinant. In particular, what is the second homology group of such ...
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### Amending flawed "proof" that homology groups are zero

I am trying to prove a certain statement that seems true based on computational data, and there is a nice argument that proves it, assuming all cycles are the simplest ones (e.g., when the only 1-...
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My question is related to this one. I thought mine could be very elementary but I'm not sure how to look into it. Let $J$ be an ideal of $R=k[\mathbf{x}]$ where $\mathbf{x}=\{x_1,\dots,x_n\}$. Let $\{... 18 votes 2 answers 569 views ### Integral homology classes that can be represented by immersed submanifolds but not embedded submanifolds Let$M$be an$m$-dimensional compact closed smooth manifold and$z\in H_n(M,\mathbb{Z})$an$n$-dimensional integral homology class, with$m>n.$Does there exist a pair of$M$and$z$so that$z$... 4 votes 0 answers 97 views ### Weaker condition for the excision axiom This comes from a question I asked on mathstackexchange (link: here) The excision axiom in homology states that if$\overline Z\subseteq\operatorname{Int}A$, then$h_n(X\setminus Z,A\setminus Z)$is ... 8 votes 1 answer 183 views ### Integral homology classes of which no multiples admit embedded representatives with trivial normal bundle Let$M$be a closed smooth manifold of dimension$n$and$z\in H_l(M,\mathbb{Z})$a$k$-dimensional integral homology class. Theorem II.4 of Thom's classical 1954 paper states that for$l< n/2$or$...
I heard something about the triple intersection number $\text{mod}(2)$ (but maybe also $\text{mod}(n)$) of a surface in an orientable three-manifold but I couldn't find a precise definition. My guess ...