nikkkom said:
Well, I don't understand how it is possible: looks like any combination of gluons alone will have an overall non-neutral color charge, and therefore such particle can not be isolated.
Properly understanding this requires group theory, but I've had a go at explaining it below:
When we say something "has neutral color charge," what we actual mean is "doesn't change under any color gauge transformations." What are color gauge transformations? Well, one example of a color gauge transformation would be the transformation: red becomes green, antired becomes antigreen, green becomes minus red, antigreen becomes minus antired. This is essentially switching the "red" and "green" color labels, so that whatever you called a red quark before you now call a green quark, or whatever you called a blue-antigreen gluon before you now call a blue-antired gluon (there's also that funny minus sign). You can switch any two color labels in this way as a color gauge transformation, or you could even do a cyclic relabeling like: red->green, green->blue, blue->red, and same for the anticolors. Actually, there are infinitely many more exotic gauge transformations, like ones that turn red into (red + green)/sqrt(2) and green into (red - green)/sqrt(2), or transformations that behave differently at different points in space and time.
So for example, a red quark is not color neutral because e.g. if you do the above gauge transformations it turns into a green quark. So this matches up with our intuition that quarks "carry color charge."
Now, you can verify that none of the gluons are invariant under *all* color gauge transformations. For example, (red-antired + blue-antiblue - 2*green-antigreen) is invariant under switching red and blue, but changes if you switch red and green. So the gluons aren't "color neutral" and this is why gluons can emit and absorb other gluons. You can also verify that the nonexistent gluon (red-antired + blue-antiblue + green-antigreen) actually *would* be invariant under all color transformations, and this is basically why it doesn't exist.
Before getting to glueballs, let's talk about mesons and baryons. Mesons consist of a quark and an anti-quark. The quark has a color and the antiquark has an anticolor. The color state of a meson is denoted (red-antired + blue-antiblue + green-antigreen). The meaning of this is basically "if you pull apart a meson, 1/3 of the time you will find a red quark and an antired antiquark, 1/3 of the time you will find a blue quark and an antiblue antiquark, and 1/3 of the time you will find a green quark and an antigreen antiquark." We can't actually pull apart mesons to look at the quarks inside, but this is the general idea. Note that this is the same color state as the nonexistent gluon--it's color neutral. It has to be, because as you know all particles that can exist in isolation must be color neutral.
Next, baryons. Baryons, e.g. the proton, consist of three quarks, and each quark has a color. Baryons have the color state (rgb - rbg + gbr - grb + brg - bgr), where I've abbreviated r = red, etc. The meaning of this is that if you "pulled apart" the proton, 1/6 of the time the first quark would be red, the second green, and the third blue, etc. This state is also unchanged if you do any color transformation. For example, let's try the one that switches red and green. It takes r to g and g to -r. So
(rgb - rbg + gbr - grb + brg - bgr) goes to (g(-r)b - gb(-r) + (-r)bg - (-r)gb + bg(-r) - b(-r)g)
Pulling out the minus signs to the front of each term this is
(-grb + gbr - rbg + rgb - bgr + brg) and you can verify that this is just a rearrangement of the original state, so it didn't actually change. This is actually the only color-invariant state of three quarks, so it must be the color state of all baryons.
So, what about glueballs? Let me pretend for simplicity that there are only two colors, r and g. Then there are three gluons, denoted ##(r \bar{g} + g \bar{r}), i(r \bar{g} - g \bar{r}), (r \bar{r} - g \bar{g})##. Here the overbars denote anticolors. Let's call these three gluons types 1, 2, and 3.
Then a color-invariant state of two gluons might be denoted (11 + 22 + 33). The meaning here is that if you "pulled apart" the glueball, 1/3 of the time you'd get two type 1 gluons, 1/3 of the time two type 2 gluons, and 1/3 of the time two type 3 gluons.
Is the 11 + 22 + 33 state color invariant? With only two colors in play, the only simple color transformation is the one that takes r to g and g to -r. You can verify that this transformation turns a type 1 gluon into a type 2 and a type 2 into a type 1 (up to signs and factors of i which I'm not going to work out). It leaves the type 3 gluon unchanged (though this gluon still isn't color neutral because there are other more exotic color transformations under which it changes). So our simple transformation takes (11 + 22 + 33) into (22 + 11 + 33) which is of course the same state. The state is indeed color invariant (though really we'd have to prove this for all the more exotic possible color transformations).
So that's a long but still shallow introduction to the meaning of "color neutral" and how a glueball can be color neutral. To get the full story, learn group theory :)