# Kinetic energy of incompressible fluid as quadratic form on tangent space.

I have started working through "Topological Methods in Hydrodynamics" by V.I. Arnold and I am confused by the following located on page 2:

"The kinetic energy of an [incompressible] fluid [with density 1] is...

\begin{align*} E = \frac{1}{2} \int_{M} v^{2} dx \end{align*}

where v is the velocity field of the fluid: $$v(x,t) = \frac{\partial}{\partial t} g^t(y), x = g^t(y)$$ (y is the initial position of the particle whose position is x at the moment t.

Suppose that the configuration $$g$$ has velocity $$\dot{g}$$. The vector $$\dot{g}$$ belongs to the tangent space $$T_{g}G$$ of the [Lie] group $$G = SDiff(M)$$. The kinetic energy is a quadratic form on this vector space of velocities"

Some definitions of the objects used:

• M is the manifold that the fluid lives in

• SDiff(M) is the connected component of the Lie group of diffeomorphisms: $$M \rightarrow M$$ that preserve the volume form (i.e $$f^{*}w = w$$, where $$f^{*}w$$ is the pullback of $$w$$ by $$f \in SDiff(M)$$ where $$w$$ is the volume form).

• x and y are elements of M

My confusion lies with the statement: "The vector $$\dot{g}$$ belongs to the tangent space $$T_{g}G$$ of the [Lie] group $$G = SDiff(M)$$"

I fail to see how $$v = \frac{\partial}{\partial t} g^t(y)$$ satisfies the definition of a tangent vector that I was taught:

My def:

Let $$\gamma$$ be a curve: \begin{align*} R & \rightarrow SDiff(M)\\ t & \mapsto g^t \end{align*}

with $$\gamma (0) = g$$

Then $$D_{\gamma}$$ is the derivation of $$\gamma$$ at $$g$$ defined by:

$$D_{\gamma}(h) = (h \circ \gamma)'(0)$$

where $$h\in C^{\infty}(M)$$

These $$D_{\gamma}$$ are the elements of $$T_{g}G$$ where $$G = SDiff(M)$$

I do not see how to interpret $$\frac{\partial}{\partial t} g^t(y)$$ as this type of object. For one thing $$g^t$$ is a map $$M \rightarrow M$$ and I am unsure of how to take the partial derivative of such a thing. Furthermore I am not sure of how I can interpret $$\frac{\partial}{\partial t} g^t(y)$$ as a map from $$C^{\infty}(M)$$ to $$R$$.