Symmetric bilinear form that is also positive definite, i.e.:

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Every vector space is defined over a field.

E.g. in ${\mathbb{R}}^3$, the underlying field is $\mathbb{R}$, the real numbers. And in ${\mathbb{C}}^2$ the underlying field is $\mathbb{C}$, the complex numbers.

Any field can be used, including finite field. But the underlying thing has to be a field, because the definitions of a vector need all field properties to hold to make sense.

Elements of the underlying field of a vector space are known as scalar.

Group of even permutations.

Note that odd permutations don't form a subgroup of the symmetric group like the even permutations do, because the composition of two odd permutations is an even permutation.

en.wikipedia.org/w/index.php?title=Mathieu_group&oldid=1034060469#Multiply_transitive_groups is a nice characterization of 4 of the Mathieu groups.

You select a generating set of a group, and then you name every node with them, and you specify:

Not unique: different generating sets lead to different graphs, see e.g. two possible en.wikipedia.org/w/index.php?title=Cayley_graph&oldid=1028775401#Examples for the

Ultimate explanation: math.stackexchange.com/questions/776039/intuition-behind-normal-subgroups/3732426#3732426

Only normal subgroups can be used to form quotient groups: their key definition is that they plus their cosets form a group.

Intuition:

One key intuition is that "a normal subgroup is the kernel" of a group homomorphism, and the normal subgroup plus cosets are isomorphic to the image of the isomorphism, which is what the fundamental theorem on homomorphisms says.

Therefore "there aren't that many group homomorphism", and a normal subgroup it is a concrete and natural way to uniquely represent that homomorphism.

The best way to think about the, is to always think first: what is the homomorphism? And then work out everything else from there.

A measurable function defined on a closed interval is square integrable (and therefore in $L^2$) if and only if Fourier series converges in $L^2$ norm the function:

This is the standard model.

Ciro Santilli's preferred visualization of the real projective plane is a small variant of the standard "lines through origin in ${\mathbb{R}}^3$".

Take a open half sphere e.g. a sphere but only the points with $z>0$.

Each point in the half sphere identifies a unique line through the origin.

Then, the only lines missing are the lines in the x-y plane itself.

For those sphere points in the circle on the x-y plane, you should think of them as magic poins that are identified with the corresponding antipodal point, also on the x-y, but on the other side of the origin. So basically you you can teleport from one of those to the other side, and you are still in the same point.

Ciro likes this model because then all the magic is confined just to the $z=0$ part of the model, and everything else looks exactly like the sphere.

It is useful to contrast this with the sphere itself. In the sphere, all points in the circle $z=0$ are the same point. But this is not the case for the projective plane. You cannot instantly go to any other point on the $z=0$ by just moving a little bit, you have to walk around that circle.

www.youtube.com/watch?v=tq7sb3toTww&list=PLxBAVPVHJPcrNrcEBKbqC_ykiVqfxZgNl&index=19 mentions that it is a bit like a dot product but for a tangent vector to a manifold: it measures how much that vector derives along a given direction.

In plain English: the space has no visible holes. If you start walking less and less on each step, you always converge to something that also falls in the space.

One notable example where completeness matters: Lebesgue integral of $L^p$ is complete but Riemann isn't.

When a disconnected space is made up of several smaller connected spaces, then each smaller component is called a "connected component" of the larger space.

See for example the