In my classical mechanics class, my professor asked if Legendre transformations form a group, and in my little knowledge about groups, I know that a transformation group consists of a set of transformations and compositions of transformation as the group operation. The first "problem" I encounter was that, the way I see it and I don't know if it is correct, is that there's only "one" legendre transformation, so my first thought was that this set would consist only of one element, and knowning that if I apply two times the legendre transformation to a function, it gives me the same function, It would be its own inverse element. But then I have problems thinking about the identity transformation, so maybe it could form a group if we consider the set of the legendre transformation plus the identity transformation. As i said before, I'm not an expert in group theory, I just started learning about groups, so if i said something incorrect, I would be grateful if you correct me. Thank you.
1 Answer
Think about a system with the so-called $C_2$ rotational axis. That is a system such that a rotation by an angle $\pi$ around that axis corresponds to a configuration indistinguishable from the original.
The set of rotations around such $C_2$ axis has the algebraic structure of a group of two elements, The identity $E$, i.e., all the equivalent rotations by $2 \pi n$, with $n$ integer, and a $R_2$ rotation, i.e., all the equivalent rotations by $\pi n$, with $n$ odd integer.
In the case of Legendre transforms, one has to:
- explicitly characterize the set elements on which the transformations apply and the set of possible transformations (this step may be non-trivial, depending on the exact definition of Legendre transform, and the corresponding domain, one is using);
- identify the composition of transformation as the sequential application of Legendre transforms;
- identify the neutral element (i.e., adding the no-transform to the definition of Legendre transforms);
- proving the existence of the inverse of each element (transform).
And that's all.