Ciro Santilli @cirosantilli  40

Intuition, please? Example? mathoverflow.net/questions/278641/intuition-for-symplectic-groups The key motivation seems to be related to Hamiltonian mechanics. The two arguments of the bilinear form correspond to each set of variables in Hamiltonian mechanics: the generalized positions and generalized momentums, which appear in the same number each.

Seems to be set of matrices that preserve a skew-symmetric bilinear form, which is comparable to the orthogonal group, which preserves a symmetric bilinear form. More precisely, the orthogonal group has:and its generalization the indefinite orthogonal group has:where S is symmetric. So for the symplectic group we have matrices Y such as:where A is antisymmetric. This is explained at: www.ucl.ac.uk/~ucahad0/7302_handout_13.pdf They also explain there that unlike as in the analogous orthogonal group, that definition ends up excluding determinant -1 automatically.

Therefore, just like the special orthogonal group, the symplectic group is also a subgroup of the special linear group.

All non-clade groups are evil. All non-clade terms must be forgotten. Some notable ones:

It is those early splinter groups that make it so clear that religion is fake. We believed in Judaism. Now we believe in Jesus. Oh, but wait, actually Judaism is now semi fake, and we don't live by Jewish law anymore.

That self blame, imminent end of the world and hatred for tax collectors stuff is a bit over the top, you should chill man.

But please keep the parables coming.

Ciro Santilli had to see this in a few separate places, until he underestood: that little pictur emust be a thing! Examples:

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The change of basis matrix $C$ is the matrix that allows us to express the new basis in an old basis:

Mnemonic is as follows: consider we have an initial basis $(x_{old},y_{old})$. Now, we define the new basis in terms of the old basis, e.g.:which can be written in matrix form as:and so if we set:we have:

The usual question then is: given a vector in the new basis, how do we represent it in the old basis?

The answer is that we simply have to calculate the matrix inverse of $M$:

That $M^{-1}$ is the matrix inverse.

When we have a symmetric matrix, a change of basis keeps symmetry iff it is done by an orthogonal matrix, in which case:

Examples: