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 $R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.