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Probably, the easiest method is this (at least in characteristic zero, which I will assume henceforth): Suppose that $V\subset \mathbb{A}^n_k$ is the set of zeros of a polynomial ideal $I\subset k[x^ …

A little algebra shows that, for $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$ on the sphere, this surface is an irreducible algebraic surface of degree 5 that is singular at $A$, $B$, and $C$ but is othe …

There are also the limaçons, which are rational curves of degree 4 with one node and two cusps: There is a $1$-parameter family of these up to projective equivalence (the parameter is $a\not= \pm1,\p …

Here's an example: $M^8 = S^1\times S^7$. This manifold is parallelizable, so it has an $\mathrm{SU}(4)$-structure. It is diffeomorphic to the quotient of $\mathbb{C}^4\setminus\{0\}$ divided by th …

Maybe this is something like what you have in mind: Let $V$ be a (complex) vector space of dimension $3$. It is easy to show that a generic subspace $P\subset S^2(V^*)$ of dimension $4$ can be writt …

Caution: I was thinking over the calculation that led to my proposed answer below, and I realized that I had neglected a term that I haven't actually justified as being zero, so now I'm less sure tha …

The classical proof via differential geometry goes like this: Suppose that the surface in $\mathbb{R}^3$ is smooth and parametrize it locally in the form $X(s,t)$ where the two rulings are defined …

That depends on what you mean by 'explicit'. For example, in Some examples of special Lagrangian tori (Adv. Theor. Math. Phys., vol. 3 no. 1 (1999), pp. 83–90, also available at arXiv:math/9902076), …

As we know, the projective hypersurface in $\mathbb{P}^n$ defined by a homogeneous polynomial equation $$ F(x^0,\ldots,x^n)=0 $$ of degree $m$ is nonsingular if $x=0$ is the only solution to the equat …

This is a trick question, right? It's not true when $m=2$ because, then $\mathrm{SO}(m{-}1)=\mathrm{SO}(1)$ is trivial, so that all polynomials on $2$-by-$2$ symmetric matrices are invariants, and th …

(N.B.: I have modified this answer to take into account the comments below about the case of characteristic $2$, when in fact, the discriminant is the square of an irreducible polynomial.) The discri …

NB: This answer is directed to the questions about the real case, not the complex case, which was already treated by Francesco. Added 5 July 2021: Because of some questions I have received over the …

The Jordan algebra is power associative, so the answer is $$ COM(X) = X^2 - tr(X) X + \sigma_2(X) I, $$ where $\sigma_2(X) = \tfrac12\bigl(tr(X)^2 - tr(X^2)\bigr)$. From another point of view, thi …

I'm not sure what you mean by 'characterize'. For example, here is a trivial characterization: Let $C$ be any Riemann surface, with meromorphic functions $a$ and $b$ with $da\not=0$. Then let $(x,y) …

Taking this as two questions, the second being more interesting than the first, I can at least answer the first (less interesting) question. The answer is 'Yes, such maps $\phi:\mathbb{C}^2\to\mathbb …