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The ordinary Topological $K$ theory defined by Atiyah and Hirzebruch is a generalized cohomology theory (see wikipedia).There is the Bott spectrum associated to this generalized cohomology theory.Using that spectrum,we could surely produce a generalized homology theory.I call this Topological $K$-homology.

While,i noticed that there are "many" $K$-homologies in the literature:

Analytic $K$-homology

Geometric $K$-homology

$KK$-theory

Is there a good survey paper on these "various" $K$-homology theories and their relationship?

there is a similar question in the equivariant case.Atiyah and Segal gave the definition of equivariant $K$-cohomology theory.

By constructing a $G$-spectrum,and then taking smash product with $G$ space,then take the stable $G$-homotopy groups,we could define generalized equivariant homology theory,in particular,equivariant $K$ homology(see Carlsson's paper"a survey of equivariant stable homotopy theory" ).

Another equivariant homology theory appeared in the literature is the Bredon type equivariant homology,where we are given a functor from orbit category Or$G$ to abelian groups,using that as "coefficient system",we could produce an "equivariant homology theory".

I guess there are some other equivariant $K$-homologies.say, index theorists usually use the notation $KO^G_i(M)$ for closed $G$-manifold $M$, which is the home of some signature operator classes.

what's the relationship between these equivariant $K$-homology theories? What's the correct reference for $KO^G_i(M)$ mentioned above?

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I would say that there are really only two definitions of K-homology commonly used in the literature (apart from the naive definition via the Bott spectrum): "analytic K-homology" and "geometric K-homology". KK theory is a bivariant theory which includes topological K-theory as the special case $KK(\mathbb{C},C(X))$ and analytic K-homology as the special case $KK(C(X), \mathbb{C})$. (Perhaps one could think of E theory as yet a third definition.)

The equivalence between geometric and analytic K-homology is simple enough. A cycle in the geometric K-homology group of a space $X$ consists of a compact Spin$^c$ manifold $M$, a Hermitian vector bundle $E$ over $M$, and a continuous map $\phi \colon M \to X$. A cycle in the analytic K-homology group of $X$ consists of a Hilbert space $H$, a representation of the C*-algebra $C(X)$ on $H$, and a bounded Fredholm operator on $H$ which is compatible with the representation. (Of course the challenge in both cases is to get the relations between cycles right.) The equivalence between the geometric and analytic models is as follows: given a geometric cycle $(M,E,\phi)$, let $H$ be the Hilbert space of $L^2$ sections of $E$ (which comes naturally equipped with a representation $\rho$ of $C(M)$), let $D_E$ be the Spin$^c$ Dirac operator on $M$ twisted by $E$, and form the bounded Fredholm operator $F = D/\sqrt{1 + D^2}$ via the functional calculus. Then $(H,\rho,F)$ is a cycle in the analytic K-homology of $M$, and by functoriality it pushes forward along $\phi$ to an analytic K-cycle for $X$.

This construction defines a map from geometric to analytic K-homology, and with a bit of effort one can show that it is an isomorphism. A good reference for the proof (including some remarks about the $KO$ story) is here: http://arxiv.org/pdf/math/0701484v4.pdf

The equivariant case is a little bit trickier, because I don't think there is universal agreement on the right definition of equivariant K-homology (unless maybe the group is compact). A good reference to get going on the analytic side, at least, is Blackadar's textbook "K-theory for Operator Algebras", which includes what you want as a special case of equivariant real KK theory.

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    $\begingroup$ @მამუკაჯიბლაძე The idea is to replace the unbounded Fredholm operator $D$ with a bounded operator which has the same index, and any operator of the form $f(D)$ where $f(t)$ is a smooth odd function on $\mathbb{R}$ which satisfies $f(t) > 0$ for $t > 0$ and $f(t) \to 1$ as $t \to \infty$ will do the job. This is explained in detail in chapter 10 of Higson-Roe. $\endgroup$
    – Paul Siegel
    Commented Apr 26, 2015 at 19:32
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    $\begingroup$ (In fact it would be more elegant to build a version of the theory in which the generators are unbounded operators since Dirac operators are supposed to give the fundamental class in K-homology, but I don't think anybody knows what the relations would look like. For instance two unbounded operators could give the same K-homology class even if their domains intersect trivially.) $\endgroup$
    – Paul Siegel
    Commented Apr 26, 2015 at 19:35
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    $\begingroup$ A possible first step towards this might be found in Hilsum's notion of bordism of unbounded KK-cycles---nothing to do with cobordism of bounded cycles---which has recently been developed further by Deeley--Goffeng--Mesland: arxiv.org/abs/1503.07398 $\endgroup$
    – Branimir Ćaćić
    Commented Apr 26, 2015 at 19:47
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    $\begingroup$ @user2015 The relationship between analytic K-homology and the Bott spectrum comes down to the fact that the space of Fredholm operators on a separable Hilbert space represents the K-theory functor. One can then build a model for the K-theory spectrum by looking at spaces of Fredholm operators which anti-commute with Clifford generators (in the spirit of the multigrading structure in Kasparov's theory). This was worked out by Atiyah and Singer in this paper: maths.ed.ac.uk/~aar/papers/askew.pdf $\endgroup$
    – Paul Siegel
    Commented Apr 27, 2015 at 19:35
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    $\begingroup$ @user2015 If you're referring to the Baum-Connes assembly map, you can formulate the map in either the analytic or geometric model of K-homology, though I suppose the analytic model is typically used. Some of the various different approaches are formulated in this paper: arxiv.org/pdf/0907.2066v1.pdf, a follow-up to the paper I linked to in my answer. (It is also useful to formulate the assembly map using the E-theory picture of K-homology; I believe it is this model that was used to prove the Baum-Connes conjecture for A-T-menable groups.) $\endgroup$
    – Paul Siegel
    Commented Apr 27, 2015 at 19:44

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