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So we know that the lead derivative of omega is zero. That's if and only if.

What happens? You take omega, you plug in x to the first place, you get a one form, you complete D of that.

If this is equal to zero, I can even perhaps give you a proof. So how does one prove it?

The trick is that L of x can be expressed as either composition of Ix with D and plus the other composition D with Ix.

And if you apply it to omega, then this term disappears because omega is closed. So this is the only term that remains.

So in fact, this expression is equal to that expression.

Okay.

And now, you see, we have now closed one form, so sometimes it happens that this closed one form is differential of some function.

So if Ix omega is equal to D of some function, so H, then we say that H is the Hamiltonian for X.

And what you can also notice is that H is unique up to a constant.

Let's say if M is connected.

And now, what happens if you have an entirely algebra acting on your simple acting manifolds?

So suppose that G is a real algebra and G acts on M.

And we suppose that this action preserves the simplistic form. So the derivative of X of omega with respect to X is equal to zero for all X and G.

Then we can try to find a Hamiltonian for each of those vector fields.

So suppose that we manage to do it for basis of G, then we have it for every element of G.

And so we can certainly suppose Hx is a Hamiltonian for X.

And we can suppose that this H depends linearly on X. So H of Ax plus By is equal to H something like this.

If you have something like that, then actually we can combine these things.

We can say that this is equivalent to having a map mu which goes from M to the dual of the Lie algebra.

So what is the relation between mu and those Hx? Hx is equal to the contraction between mu and X.

That's for every X in G.

So just having something like that is the same as having a map.

And what can we say about this map? We know, for example, that if you take the Poisson bracket of Hx and Hy,

then by the property of Poisson brackets, we know that this thing is actually the Hamiltonian for the commutator of X and Y.

So we know that this is equal to H of the commutator. But there might be a plus. What is the problem?

See, a Hamiltonian is unique up to a constant. So this is A Hamiltonian. This is also A Hamiltonian for a bracket of X and Y.

They might not be equal. It might happen that there is some additional term.

Cxy, this is just a constant, a real number. So the C goes to G times G to R.

And it satisfies some properties, so it's a cos cycle.

And if C is simply equal to 0, then we say that mu is a moment map.

So let me give you an equivalent definition of exactly what now happened. I will rephrase it slightly differently,

because when we modify this definition, it will be more convenient like that.

So I can say that a moment map, or let me phrase it like this, a Hamiltonian G manifold is a simplex manifold.

It's an omega with an action of G on M preserving omega.

And a map mu which goes from M to the dual of the Lie algebra, such that...

What does it work? This is also E.

There is one condition which will say that the corresponding Hx is a Hamiltonian.

So the first condition is that Ix to omega is equal to this contraction between mu and x.

This form, x and G.

And the second condition is going to be equivalent to the vanishing of this cos cycle.

So the second condition says that mu is G equivalent

with respect to the co-adjoint action of G on G star.

So let me take some time to explain what is written here.

By the way, maybe... Do you have any questions, for example?

Anyway, you might try to suggest me by some movements, if you know everything and I should go faster,

but more importantly, if I go too fast, then I should slow down.

Could you say again how this object in the first condition is defined on the right hand side?

Mu is something that has values in G star.