@Semiclassical My understanding is that Hodge theory circles around ideas involving the Hodge theorem, which IIRC says on a Riemannian manifold any de Rham cohomology class admits a harmonic closed form which is in that class (aka a harmonic representative) .
That involves the word "cohomology", but I think the point is geometric there. de Rham cohomology has more structure than an arbitrary cohomology theory.
the construction of singular cohomology is not too different from simplicial. it's easier to handle, too, because you don't have to deal with combinatorial arguments
$$(\sum_{i=1}^{n+1}x_i^2)\cdot (\sum_{i=1}^{n+1}y_i^2)$$ $$=(x_{n+1}^2+\sum_{i=1}^{n}x_i^2)\cdot (y_{n+1}^2+\sum_{i=1}^{n}y_i^2)$$ $$=(\sum_{i=1}^{n}x_i^2)\cdot (\sum_{i=1}^{n}y_i^2)+x_{n+1}^2\cdot (\sum_{i=1}^{n}y_i^2)+y_{n+1}^2\cdot (\sum_{i=1}^{n}x_i^2)+x_{n+1}^2\cdot y_{n+1}^2$$ We now estimate this using the assumption $$\geq (\sum_{i=1}^{n}x_iy_i)^2+x_{n+1}^2\cdot (\sum_{i=1}^{n}y_i^2)+y_{n+1}^2\cdot (\sum_{i=1}^{n}x_i^2)+x_{n+1}^2\cdot y_{n+1}^2$$
@DHMO if I'm not using induction, would an argument be: a summation of squares is a summation of positive terms. The squaring of an arbitary summation is therefore always smaller then the former. (cauchy schwarz inequality)
Affine geometry is exactly the same as vector geometry, but without a canonical choice of the origin. So it's everything what happens in vector geometry, modulo translations.
@DHMO First division by zero divisor algebra that does not contain a null semigroup NB and yes, with only 5 elements, you cannot make it associative since the bright red entries are locked out and they are the ones that break associativity. Also if both division by zero divisors and division by zero are present, the structure collapses into the 4 element structure (which we have previously showed to always contain a null semigroup).
