Let $X$ be an irreducible variety of dimension $d$. A point $p\in X$ is nonsingular, or smooth, whenwhere $T_{p}X$ is the Zariski tangent space; it is a singular point when $\dim T_{p}X>d$. Equivalently, the local ring ${\mathcal{O}}_{X,p}$ is regular exactly at a nonsingular point.

For an irreducible affine variety $X\subseteq {\mathbb{A}}^N$ over a perfect field, the Jacobian criterion expresses the singular locus by the vanishing of the relevant Jacobian minors, so it is Zariski closed. At a point where the tangent-space dimension is minimal it equals $d$; equivalently, some $(N-d)$-rowed Jacobian minor is nonzero. Its nonvanishing locus is therefore a nonempty Zariski-open set contained in the smooth locus. Every nonempty open subset of an irreducible topological space is dense, proving the density of the smooth locus.

Assume the ground field has characteristic other than two. Forthe partial derivatives are $2X_0,\dots ,2X_{n-1},0$. They vanish simultaneously at the unique projective pointThus this point is the singular locus of the projective quadric cone.

For varieties $X,Y$,If $Y$ is smooth of dimension $m$, the product point $(x,y)$ is singular exactly when $x$ is singular. HenceandThis is the singular locus of a product with a smooth variety.

To construct the requested examples, put $r=n-k>0$. If $r\ge 2$, let $Q_r\subseteq {\mathbb{P}}^{r+1}$ be an irreducible quadric cone of dimension $r$ with one singular vertex. If $r=1$, use instead the irreducible cuspidal cubicwhose only singular point is $[0:0:1]$. Let $W$ denote the chosen $r$-dimensional variety and embedin projective space by the Segre embedding. It is irreducible and has dimension $r+k=n$, while its singular locus is the vertex or cusp point times ${\mathbb{P}}^k$, which is nonempty and has dimension exactly $k$. This is the projective variety with a prescribed-dimensional singular locus construction.

Finally, suppose that the irreducible plane curve $C\subseteq {\mathbb{P}}^2$ were smooth of degree $d$. Since it is birational to a smooth projective curve of genus two, its geometric genus would be two. But the genus of a smooth plane curve isand no integer $d$ makes this number equal to two. Therefore $C$ cannot be smooth and must contain a singular point.

Use the conventionThe Riemann-Lebesgue lemma states that if $f\in L^1({\mathbb{R}}^n)$, then $\hat{f}$ is continuous andContinuity follows from the dominated convergence theorem, since $e^{-ix\cdot {\xi }_j}\to e^{-ix\cdot \xi }$ and the integrands are dominated by $|f(x)|$.

For the decay, first take $g\in C_c^1({\mathbb{R}}^n)$. Choose a coordinate $j$ for which $|{\xi }_j|\ge |\xi |/\sqrt{n}$. Integration by parts givesand henceSince $C_c^1({\mathbb{R}}^n)$ is dense in $L^1({\mathbb{R}}^n)$, choose $g$ with $\Vert f-g{\Vert }_1<\epsilon$. The elementary Fourier bound givesso the decay for $g$ implies the decay for $f$.

The Parseval identity says that for $f,g\in L^2({\mathbb{R}}^n)$, with their Fourier transforms defined by the Plancherel theorem,In particular,

For the given radial function, as $|x|\to 0$,while as $|x|\to \infty$,Using polar coordinates, local integrability of $|f{|}^q$ is therefore determined byand integrability at infinity byThe assumptions $2a>-n$ and $b>n$ imply both $f\in L^2({\mathbb{R}}^n)$ and $f\in L^1({\mathbb{R}}^n)$. The Riemann--Lebesgue lemma and the direct estimate $|\hat{f}|\le \Vert f{\Vert }_1$ give $\hat{f}\in L^{\infty }$, while Parseval gives $\hat{f}\in L^2$. Finally, for every $2\le p<\infty$,Thus

Suppose first that $F:D^n\to X$ extends $f$. Regarding $S^{n-1}$ as the boundary of the unit disc,is a homotopy from $f$ to the constant map with value $F(0)$. Thus $f$ is null-homotopic. Conversely, if $H:S^{n-1}\times [0,1]\to X$ is a homotopy from $f$ to a constant, then $H$ is constant on $S^{n-1}\times \{1\}$ and hence descends to the quotientwhich is the cone on $S^{n-1}$ and is homeomorphic to $D^n$. The descended map extends $f$. This proves the extension-null-homotopy criterion for a sphere.

A universal cover of $X$ is a covering map $p:\tilde{X}\to X$ whose total space is path-connected and simply connected. Let $p_1:({\tilde{X}}_1,{\tilde{x}}_1)\to (X,x)$ and $p_2:({\tilde{X}}_2,{\tilde{x}}_2)\to (X,x)$ be two universal covers. The lifting criterion for a covering space applies becauseso $p_1$ has a based lift $h:{\tilde{X}}_1\to {\tilde{X}}_2$ satisfying $p_{2}h=p_1$. Similarly there is a based lift $k:{\tilde{X}}_2\to {\tilde{X}}_1$ of $p_2$. Both $kh$ and the identity are lifts of $p_1$ that agree at the base point, so uniqueness of lifts gives $kh=1$. Likewise $hk=1$. Hence $h$ is a homeomorphism. This is the uniqueness of a universal covering space.

Now let $p:\tilde{X}\to X$ be universal with $\tilde{X}$ contractible. If an extension $F:|K|\to X$ exists, functoriality of the fundamental group givesThus the required factorization holds with $\Phi =F_{\ast }$.

Conversely, suppose $f_{\ast }=\Phi i_{\ast }$. We extend over the simplices of $K$ by dimension. The boundary of every 2-simplex is a loop in $|K^1|$ that becomes null-homotopic in $|K|$, so its class lies in $\ker i_{\ast }$. The factorization givesHence $f{|}_{\partial \sigma }$ is null-homotopic in $X$ and extends over the disc $|\sigma |$ by the first part. Since distinct 2-simplices meet along the already fixed 1-skeleton, these extensions combine to a map on $|K^2|$.

Inductively suppose the map is defined on $|K^{m-1}|$ for $m\ge 3$. For an $m$-simplex $\sigma$, its boundary is $S^{m-1}$, which is simply connected. The boundary map therefore lifts through $p$ by the covering-space lifting criterion. Its lift into the contractible space $\tilde{X}$ is null-homotopic, so the boundary map itself is null-homotopic and extends over $|\sigma |$. Extending over every $m$-simplex and then every skeleton constructs $F:|K|\to X$. The weak topology on a simplicial complex makes the cellwise map continuous. This proves the extension criterion into a space with contractible universal cover.

The Lagrange theorem for polynomial congruences says that if $p$ is prime and $f\in \mathbb{Z}[X]$ has degree $d$ with at least one coefficient not divisible by $p$, thenhas at most $d$ incongruent solutions modulo $p$.

The Chinese remainder theorem states that for pairwise coprime $m_1,\dots ,m_r$, every systemhas exactly one solution modulo $M={\prod }_i{m}_i$. For two moduli, choose $u,v$ with $um+vn=1$ by Bezout identity. Thenis congruent to $a$ modulo $m$ and to $b$ modulo $n$. If $x,y$ are two solutions, both $m$ and $n$ divide $x-y$; coprimality makes $mn$ divide $x-y$. This proves existence and uniqueness for two factors, and induction proves the general statement.

Nowand $12^3+1=1729$, so $x=12$ is a solution. For $1\le x<12$, one has $0<x^3+1<1729$, so no such positive integer can satisfy the congruence. Hence the smallest isModulo $7,13,19$, the roots are respectivelyEach list has three elements, and the Chinese remainder theorem combines the choices independently. Therefore the number of solutions with $1\le x\le 1729$ is

Let $V_n$ be the space of homogeneous degree-$n$ polynomials in $x,y$. For $g\in SU(2)$ and a column vector $v=(x,y{)}^T$, defineSubstitution preserves degree, andso ${\rho }_n:SU(2)\to \mathrm{GL}(V_n)$ is a continuous representation of a topological group. This is the homogeneous polynomial representation of SU2.

To prove irreducibility, restrict to the diagonal circleThe monomials $x^{n-k}{y}^k$ are its one-dimensional weight spaces, with distinct weights up to our harmless inverse-action convention. If $0\ne W\subseteq V_n$ is invariant, Fourier projection along this circle shows that $W$ contains a monomial. Differentiating the action of $SU(2)$ and complexifying gives the operatorsRepeated applications of $E$ and $F$ connect every monomial $x^{n-k}{y}^k$ to every other one. Hence $W$ contains the entire monomial basis, so $W=V_n$.

Every $g\in SU(2)$ is conjugate to $\mathrm{diag}(z,z^{-1})$ with $|z|=1$. Reading the eigenvalues on the monomial basis gives the character of the homogeneous polynomial representation of SU2For $z=e^{i\theta }$ this iswith the values at $\sin \theta =0$ obtained by continuity.

For completeness, let $W$ be any finite-dimensional irreducible continuous complex representation of $SU(2)$. Its restriction to the diagonal circle splits into integral weight spaces. Choose a vector of largest weight $n$. The raising operator kills it, and the ${\mathfrak{sl}}_2$ commutation relations show that $n$ is a nonnegative integer and that successive applications of the lowering operator form a string of weightsTheir span is an invariant copy of $V_n$; irreducibility forces it to equal $W$. Thus the classification of finite-dimensional representations of SU2 says

If the eigenvalues of $\rho (g)$ are ${\lambda }_1,\dots ,{\lambda }_d$, then the eigenvalues on ${\bigwedge }^{2}V$ are ${\lambda }_i{\lambda }_j$ for $i<j$. Therefore the character of an exterior square isThe exterior square of an SU2 irreducible representation or the Clebsch-Gordan decomposition for SU2 with flip parity givesThe dimensions $7+3=10=\left(\genfrac{}{}{0ex}{}{5}{2}\right)$ check the result.

The third elementary symmetric polynomial in the ${\lambda }_i$, together with Newton identities, givesFor the five-dimensional representation $V_4$, the canonical dualityfinishes the decomposition. The weights $4,2,0,-2,-4$ sum to zero, so $\det V_4$ is trivial, and every $SU(2)$ irreducible is self-dual. Consequently