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An algebraic group is connected when its underlying Zariski topology is a connected space. An unipotent algebraic group is a linear algebraic group whose elements, in a faithful matrix realization, satisfy that $u-I$ is nilpotent. A reductive algebraic group, in the connected convention, is a connected linear algebraic group whose unipotent radical is trivial: it has no nontrivial connected normal unipotent algebraic subgroup.

Let $E={\mathrm{End}}_Q(X)$. It is a finite-dimensional associative algebra, and ${\mathrm{Aut}}_Q(X)=E^{\times }$. The latter is the nonempty open subset where the determinant on ${⨁}_i{X}_i$ is nonzero. Since the affine space $E$ is irreducible, this open subset is irreducible and hence connected. It is a linear algebraic group: within $\mathrm{GL}({⨁}_i{X}_i)$ it is cut out by the linear equations of preserving the vertex summands and commuting with the arrow maps. Thus the automorphism group is connected.

By the Krull–Schmidt theorem, write $X\cong {⨁}_{a=1}^s{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposables $M_a$. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local algebra. Its residue division algebra is $k$: every element of a finite-dimensional division algebra over the algebraically closed field has an eigenvalue for left multiplication, and subtracting that scalar gives a noninvertible element, hence zero.

The standard radical description of the endomorphism algebra of a Krull-Schmidt decomposition therefore givesConcretely, maps between nonisomorphic summands lie in the radical, while on each isotypic block one reduces all entries modulo the local radical. The off-diagonal rule follows because a composition $M_a\to M_b\to M_a$ cannot be a unit for $a\ne b$: it would split $M_a$ off the indecomposable $M_b$, forcing an isomorphism. The finite-dimensional radical description then identifies the kernel of this block reduction with $J(E)$. Put $U=1+J(E)$. The Jacobson radical of the finite-dimensional algebra is nilpotent, so every $1+j$ is a unit and is unipotent on ${⨁}_i{X}_i$. Also $U$ is closed, being the translate of the linear subspace $J(E)$, and normal, being the kernel ofThis map has an explicit algebraic section: after fixing the direct-sum decomposition, a matrix $B_a$ acts on the $m_a$ copies by $B_a\otimes 1_{M_a}$. These block actions give a subgroup $L\cong {\prod }_a{\mathrm{GL}}_{m_a}(k)$ with trivial intersection with $U$. Every unit is uniquely a product of an element of $U$ and one of $L$. HenceThis Levi decomposition of a quiver automorphism group holds in every characteristic; the section is constructed directly from the multiplicity spaces.

All vector spaces in this solution are complex, while convex combinations use real coefficients. An extreme point $x$ of a convex set $C$ is one for which $x=ty+(1-t)z$, $y,z\in C$ and $0<t<1$, forces $y=z=x$. Equivalently, $x$ is not the midpoint of two distinct points of $C$.

The Krein-Milman theorem says that every nonempty compact convex set $K$ in a Hausdorff locally convex space is the closed convex hull of its extreme points. Here is a proof. A face of a convex set $F\subseteq K$ is a convex set such that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. Consider nonempty compact faces, ordered by reverse inclusion. A chain has nonempty intersection by compactness and the finite intersection property; that intersection is again a compact face. The Zorn lemma therefore gives a minimal nonempty compact face $F$.

If $F$ contains distinct $x,y$, a continuous real linear functional $\ell$ on the underlying real locally convex space distinguishes them. Such a linear functional exists because the space is Hausdorff and locally convex, by the Hahn-Banach theorem. The maximizers of $\ell$ on $F$ form a nonempty proper compact face of $F$. A face of a face is a face of $K$, contradicting minimality. Therefore $F$ is a singleton, yielding an extreme point. The same argument inside any nonempty compact face of $K$ supplies an extreme point of $K$ lying in that face.

Let $D$ be the closed convex hull of the extreme points of $K$. It is a nonempty closed subset of compact $K$, hence compact. If $x_0\in K\setminus D$, the Hahn-Banach separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_D\ell$. Its maximizer set on $K$ is a nonempty compact face, which contains an extreme point $e$ of $K$. But $e\in D$ and $\ell (e)\ge \ell (x_0)>{\sup }_D\ell$, a contradiction. Thus $\boxed{K=\overline{\mathrm{co}}(\mathrm{ext}K)}$, proving Krein-Milman theorem.

For a nonempty compact Hausdorff space $K$, the extreme points of the dual unit ball of C(K) areThis is the permitted description without proof, with $C(K)$ denoting the complex space of continuous functions on a compact space equipped with the supremum norm. If $K$ is empty, $C(K)=\{0\}$ and the closed unit ball of its continuous dual space has the single extreme point $0$ instead.

The complex Banach–Stone theorem states that a surjective complex-linear isometric isomorphism of normed spaces $T:C(K)\to C(L)$ has the formwhere $u\in C(L)$ and $\phi :L\to K$ is a homeomorphism. Conversely, every such map is a surjective complex-linear isometric isomorphism of normed spaces for the supremum norm.

To prove this, suppose first that $K,L$ are nonempty. The Banach-space adjoint $T^{\ast }$ is a bijective isometric isomorphism of normed spaces on the continuous dual spaces, so it bijects their closed unit balls and preserves extreme points. The displayed extreme point description gives uniquelyUniqueness follows by evaluating at the constant function $1$, and then using that continuous functions separate points of a compact Hausdorff space. Evaluation gives the required formula for $Tf$, and $u=T1$ is continuous. Surjectivity of $T^{\ast }$ on extreme points proves surjectivity of $\phi$: the preimage of any ${\delta }_x$ is $\alpha {\delta }_y$ for some $y$. If $\phi (y_1)=\phi (y_2)$, every function in the range of $T$ has equal values at these two points after division by $u$; surjectivity of $T$ and separation of points force $y_1=y_2$. Thus $\phi$ is bijective.

For every $f\in C(K)$, $f\circ \phi =(Tf)/u$ is continuous. The evaluation map $K\to {\mathbb{C}}^{C(K)}$, $x\mapsto (f(x){)}_f$, is a continuous injection of a compact Hausdorff space into a Hausdorff product topology, hence a homeomorphism onto its image. Continuity of every coordinate $f\circ \phi$ proves continuity of $\phi$. A continuous bijection between compact Hausdorff spaces is a homeomorphism. Conversely the weighted-composition formula plainly preserves the supremum norm, and its inverse isIf one compact space is empty, a surjective isometric isomorphism of normed spaces forces the other to be empty, and the empty homeomorphism gives the corresponding trivial case. This completes Banach–Stone theorem.

Neither the space of sequences converging to zero $c_0$ nor $L^1[0,1]$ can be isometrically a continuous dual space of a Banach space. Indeed, every nonzero continuous dual space has a nonempty weak-star compact closed unit ball by Banach-Alaoglu theorem, and Krein-Milman theorem then guarantees an extreme point. A bijective linear isometry preserves extreme points of closed unit balls.

The closed unit ball of $c_0$ has no extreme points. Given $x\in c_0$ with $\Vert x{\Vert }_{\infty }\le 1$, choose $n$ with $|x_n|<1$ and $0<\epsilon <1-|x_n|$. The two distinct elements $x\pm \epsilon e_n$ remain in that closed unit ball and have midpoint $x$.

The closed unit ball of the complex Lp space $L^1[0,1]$ also has no extreme points. An element of norm less than one can be perturbed by a sufficiently small nonzero Lp space element in both directions. If $\Vert f{\Vert }_1=1$, the non-atomic measure $|f(t)|dt$ admits a measurable set $A$ of mass $1/2$; for example the continuous function $s\mapsto {\int }_0^s|f(t)|dt$ attains $1/2$. Put $h=f(1_A-1_{A^c})$. For $0<\epsilon <1$, the distinct functions $f\pm \epsilon h$ both have Lp normwith the two coefficients interchanged for the minus sign. Their midpoint is $f$. Thus neither proposed space is isometrically a Banach dual.

Finally, $[0,1]$ is a connected space, whereas $[0,1]\cup [2,3]$ is disconnected. They cannot be homeomorphic. By Banach–Stone theorem, $C[0,1]$ and $C([0,1]\cup [2,3])$ are not isometrically isomorphic as complex Banach spaces.

All Topological K-theory groups below are complex and are indexed modulo two using Bott periodicity. For $n\ge 1$, Complex K-theory of a sphere givesFor an even-dimensional sphere, choose a generator $u$ of Reduced topological K-theory. Its square vanishes: the Chern character sends $u$ to a top-degree cohomology class, whose square is zero, and the Chern character is injective on this torsion-free abelian group. Thus the ring answers areThe exceptional zero-dimensional sphere consists of two points: $K^0(S^0)=\mathbb{Z}\times \mathbb{Z}$ with coordinatewise multiplication and $K^{-1}(S^0)=0$. In particular its reduced generator is idempotent rather than square-zero.

For the Euler characteristic identity, induct over the cells of a finite CW complex. The starting CW complex consisting of $v$ vertices has $K^0\cong {\mathbb{Z}}^v$, $K^{-1}=0$, and $\chi =v$. If $X$ is obtained from $A$ by attaching one $d$-dimensional cell, the quotient is $S^d$, and the K-theory six-term exact sequence isThese abelian groups are finitely generated by induction. Tensoring the exact sequence with $\mathbb{Q}$ preserves exactness. The alternating sum of dimensions in a cyclic six-term exact sequence is zero, soTherefore

There is a genuine convention issue in the nilpotence assertion. With the usual definition of Reduced topological K-theory as the kernel of restriction to one basepoint, the assertion needs $X$ to be connected. For $X=S^0$ based at its first point, $(0,1)$ lies in ${\tilde{K}}^0(X)$ and satisfies $(0,1{)}^m=(0,1)$ for every $m\ge 1$. Thus it is not a nilpotent element. For an arbitrary finite CW complex, the correct assertion uses the rank map in topological K-theory on every connected component:For a connected space, $I(X)={\tilde{K}}^0(X)$.

Here is an induction proving the corrected assertion. On the vertices, $I(X)=0$. For a positive-dimensional cell attachment $A\subset X$, the ideal $J=\ker (K^0(X)\to K^0(A))$ is square-zero. Indeed, lift two elements of $J$ to relative topological K-theory $K^0(X,A)$. Their product is induced by the reduced diagonalThis map is null-homotopic, since $d<2d$ and $S^{2d}$ is $(2d-1)$-connected. The relative product in topological K-theory is therefore zero, giving $J^2=0$. If $a\in I(X)$, its restriction belongs to $I(A)$, so induction gives $(a{|}_A{)}^m=0$ for some $m$. Then $a^m\in J$ and $a^{2m}=0$. We have proved every Topological K-theory class of rank zero on every connected component is a nilpotent element, and hence the requested result for connected $X$.

Suppose for contradiction that the continuous image of a connected space $Y=f(X)$ is disconnected. Then there are disjoint nonempty sets $U,V$, open in the subspace topology on $Y$, with $Y=U\cup V$. By continuity, $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open subsets of $X$; by surjectivity both are nonempty, and their union is $X$. This is a separation of a topological space of $X$, contradicting that $X$ is a connected space. Hence $Y$ is connected.

Write $I^{+}(U)=U=I^{-}(U)$. Since every chronological future and chronological past is open, $U$ is an open set. We show its complement is also open.

Take $p\notin U$ and choose $p_{-}\ll p\ll p_{+}$ in a convex normal neighborhood. The diamond $V=I^{+}(p_{-})\cap I^{-}(p_{+})$ is a neighborhood of $p$. If $x\in V\cap U$, then $p_{-}\ll x$ and $U=I^{-}(U)$ imply $p_{-}\in U$; then $p_{-}\ll p$ and $U=I^{+}(U)$ imply $p\in U$, a contradiction. Thus $V$ lies in the complement of $U$.

Hence $U$ is nonempty, open, and closed. Since $\mathcal{M}$ is a connected space,

The polynomial Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $p$ such that ${\sup }_{z\in K}|p(z)-f(z)|<\epsilon$.

For an explicit approximation on the left semicircle $K=\{z:|z|=1,\mathrm{Re}z\le 0\}$, defineThe two power-series expansionsproduce this formula. The first is a geometric series whose ratio has modulus at most $4/\sqrt{17}<1$ on $K$. The second converges uniformly on the whole unit circle; the elementary bound $\left(\genfrac{}{}{0ex}{}{j+k}{k}\right)\le 2^{j+k}$ shows that truncating it at $k=n^2$ gives a total error tending to zero even after summing over $0\le j\le n$. Consequently $p_n\to 1/z$ uniformly on $K$. This is an explicit polynomial approximation of the reciprocal on the left semicircle.

No such sequence exists on the punctured unit circle $S=\{z:|z|=1,z\ne 1\}$. If polynomials $p_n$ converged uniformly there, they would be Uniformly Cauchy. Continuity givesso they would converge uniformly on the entire unit circle. The uniform limit theorem would force the value at the missing point to be $1$, and hence the limit would be $1/z$ on the full circle. But the Cauchy integral theorem gives $\oint p_n(z)dz=0$ for every $n$, while uniform convergence and contour integration would implya contradiction. This is the polynomial approximation obstruction on a punctured circle.

This is always true. The continuous image $f(X)$ of a connected space is connected. Its closure is also connected: if the closure were separated into disjoint nonempty relatively open sets, connectedness would put $f(X)$ inside one of them, preventing its closure from meeting the other. Since $f(X)$ is dense in $Y$, its closure is $Y$, so $Y$ is connected.

A topological space is connected when it is not the union of two disjoint nonempty open subsets. It is path-connected when every two points $x,y$ can be joined by a path, meaning a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$.

To prove that $[0,1]$ is connected, suppose it had a separation $[0,1]=U\cup V$ with $0\in U$. LetWhichever of $U$ or $V$ contains $c$, its relative openness supplies an interval about $c$ that contradicts either the definition of the supremum or the fact that all points immediately below $c$ lie in $U$. Thus no separation exists.

If $X$ is path-connected and $X=U\cup V$ were a separation, choose $u\in U$ and $v\in V$. A path from $u$ to $v$ would make $[0,1]$ the union of the disjoint nonempty open sets ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$, contradicting the connectedness just proved. Hence

Now let $X\subseteq {\mathbb{R}}^n$ be open. Every point $x\in X$ lies in an open ball contained in $X$, and an open ball is path-connected by straight line segments. It follows that every path component of $X$ is open. If there were more than one path component, one component and the union of all the others would separate $X$. Consequently a connected open subset of Euclidean space is path-connected. The converse follows from the preceding implication.

The same argument answers the locally Euclidean question affirmatively. Every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$, hence a path-connected open neighbourhood after restricting to a sufficiently small ball. Thus $X$ is locally path-connected, its path components are open, and connectedness forces there to be only one.

For the final example, putThe set $Y$ is path-connected: each vertical segment $C_n$ meets the horizontal segment $A$. The segment $B$ is also path-connected and lies in the closure of $Y$, because $(1/n,y)\to (0,y)$ for every $y\in [1/2,1]$. A connected set together with a connected subset of its closure is connected, so

It is not path-connected. The image of the first coordinate of any path in $X$ lies inbut, while the path has positive height, its first coordinate lies in the totally disconnected set $\{0\}\cup \{1/n:n\ge 1\}$. A path beginning on $B$ cannot leave $x=0$ before reaching height zero, and $X$ contains no point $(0,y)$ with $0\le y<1/2$. Hence no path joins $B$ to $Y$, and

An analytic continuation of $f_1$ from ${\mathcal{D}}_1$ to ${\mathcal{D}}_2$ is a holomorphic function $f_2$ on ${\mathcal{D}}_2$ whose restriction to the nonempty overlap ${\mathcal{D}}_1\cap {\mathcal{D}}_2$ agrees with $f_1$.

Suppose $f_2$ and ${\tilde{f}}_2$ are two such continuations. Both equal $f_1$ on ${\mathcal{D}}_1\cap {\mathcal{D}}_2$, soon that nonempty open set. Since ${\mathcal{D}}_2$ is connected, the identity theorem givesThus analytic continuation to ${\mathcal{D}}_2$ is unique.

Take the topologist's sine curveIt is closed and bounded, hence compact. The oscillating graph is connected and accumulates on every point of the vertical segment, so its closure $S$ is connected. No path joins the graph to the vertical segment, as continuity of the first coordinate and the unbounded oscillation of $\sin (1/x)$ would fail at the first time the path reaches $x=0$. Thus $S$ is not path connected.

Suppose first that the left-invariant 1-form $\alpha$ is closed. The scalar function $\alpha (l_{\xi })$ is constant for every $\xi$, because both the form and vector field are left-invariant. Cartan's magic formula therefore givesPart b says that $g\mapsto R_g^{\ast }\alpha$ is locally constant. Since $G$ is connected, it is constant, and its value at the identity is $\alpha$. Hence $R_g^{\ast }\alpha =\alpha$ for every $g$, so $\alpha$ is bi-invariant.

Conversely, if $\alpha$ is bi-invariant, then all these Lie derivatives vanish. Cartan's formula and the constancy of $\alpha (l_{\xi })$ give ${\iota }_{l_{\xi }}d\alpha =0$. The left-invariant vector fields span every tangent space, so $d\alpha =0$.

Connectedness cannot be omitted. The orthogonal group $O(2)$ has an abelian Lie algebra, so every left-invariant 1-form is closed by the Maurer-Cartan equation in a Lie-algebra basis. Conjugation by a reflection acts as $-1$ on its one-dimensional Lie algebra, so a nonzero left-invariant 1-form is not right-invariant. This is the standard obstruction recorded by the closed left-invariant 1-form criterion.

A space is connected when it is not a union of two disjoint nonempty open subsets. If $f(X)$ had such a separation, its inverse images would separate connected $X$, so continuous images of connected spaces are connected. Equivalently, a nonconstant continuous map to discrete $\{0,1\}$ records a separation, and every separation defines such a map.

If $\mathbb{R}$ were separated, points in opposite pieces and the intermediate value theorem applied to the associated $\{0,1\}$-valued map would give a contradiction. Thus $\mathbb{R}$ is connected. Its quotient by $x\sim y$ is connected because the quotient map is continuous and surjective.

Finally let $A\subset B\subset \mathrm{Cl}(A)$ and suppose $B=U\cup V$ were a separation. Connectedness puts $A$ wholly in one side, say $U$. Every point of $V$ nevertheless lies in the closure of $A$, while $V$ is relatively open and disjoint from $A$, a contradiction. Hence $B$ is connected.

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives

For a connected Riemannian manifold $(M,g)$, its Riemannian distance iswhere the infimum is over the piecewise smooth curves from $p$ to $q$. Connectedness of a smooth manifold implies path connectedness, so this set of curves is nonempty.

The Gauss lemma says that the differential of ${\exp }_p$ preserves the radial inner product: for $v,w\in T_{p}M$,Consequently radial geodesics from $p$ are orthogonal to the images of tangent vectors to spheres centred at the origin in $T_{p}M$. In a sufficiently small normal neighbourhood of $p$, this impliesevery competing curve has length at least the total variation of its radial coordinate, and the radial geodesic has that length.

The axioms $d_g(p,q)\ge 0$, symmetry, and the triangle inequality follow directly from length and concatenation. Certainly $d_g(p,p)=0$. If $q\ne p$, choose a normal ball $B_g(p,r)$ that does not contain $q$. Every curve from $p$ to $q$ first meets its boundary, and its initial part has length at least $r$ by the Gauss lemma. Hence $d_g(p,q)\ge r>0$. Thus $d_g(p,q)=0$ if and only if $p=q$.