Complexified complements in an open convex cone

Let $E$ be a finite-dimensional real vector space with the Euclidean topology supplied by linear coordinates. Let $T\subseteq E$ be nonempty, open and convex, and let $\mathcal{A}$ be a locally finite arrangement in $T$: its members are linear hyperplanes meeting $T$, and every point of $T$ has a neighbourhood meeting only finitely many members. Define the complement

$M(T,\mathcal{A})=(T\times T)\setminus {\bigcup }_{H\in \mathcal{A}}(H\times H)\mathrm{.}$

Give $T\times T$ the product topology and $M(T,\mathcal{A})$ its subspace topology: its open sets are intersections of $M(T,\mathcal{A})$ with open subsets of $T\times T$.

When $T=E$, put $E_{\mathbb{C}}=\mathbb{C}{\otimes }_{\mathbb{R}}E$ and identify $(x,y)$ with $1\otimes x+i\otimes y$, also written $x+iy$. For each $H\in \mathcal{A}$, its complexification inside $E_{\mathbb{C}}$ means $H_{\mathbb{C}}={\mathrm{span}}_{\mathbb{C}}\{1\otimes h:h\in H\}$. Under this identification the complement is $E_{\mathbb{C}}\setminus {\bigcup }_{H\in \mathcal{A}}{H}_{\mathbb{C}}$.

Suppose a group $G$ acts on $E$ by invertible real-linear maps and each $g\in G$ preserves $T$ and permutes $\mathcal{A}$. The action convention is $ex=x$ and $(gh)x=g(hx)$. The diagonal action on $T\times T$ is $g\cdot (x,y)=(gx,gy)$; its restriction to the complement is the diagonal action on $M(T,\mathcal{A})$.

A homeomorphism is a continuous bijection with continuous inverse. A covering-space action is an action by homeomorphisms for which every point has an open neighbourhood $U$ satisfying $gU\cap U=\mathrm{\varnothing }$ for every nonidentity $g$. Setwise preservation alone does not assert this condition or path connectedness. Path connectedness means that for any $p,q\in M(T,\mathcal{A})$ there is a continuous map $\gamma :[0,1]\to M(T,\mathcal{A})$ with $\gamma (0)=p$ and $\gamma (1)=q$, where $[0,1]$ has its usual subspace topology from $\mathbb{R}$.

Locally finite arrangements and relative face terminology

The tensor and direct-sum models of complexification are canonically complex-linearly isomorphic

The product set ∏_i ∈ I Xᵢ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space

Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace

Covering-space actions by disjoint translates of neighbourhoods

Complexification as ℂ⊗_ℝV with its canonical real-linear embedding

The direct-sum model V⊕ iV of a complexification

Left group actions, transitive actions, and faithful actions

Paths, path-connected spaces and path components

Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological

The arrangement convention supplies the ambient topology and the meanings of convexity, hyperplane and local finiteness. The displayed set is a subset of $T\times T$, so the product and subspace topologies determine its topology.

The tensor/direct-sum isomorphism has inverse ${\mathrm{\Psi }}_E(x,y)=1\otimes x+i\otimes y$ and forward map ${\mathrm{\Phi }}_E(1\otimes x+i\otimes y)=(x,y)$. A finite complex linear combination of $1\otimes h_j$, with $h_j\in H$, can be written as $1\otimes h+i\otimes k$ with $h,k\in H$: write each coefficient as $a_j+i{b}_j$ and set $h={\sum }_j{a}_j{h}_j$, $k={\sum }_j{b}_j{h}_j$. Conversely every such expression belongs to the stated complex span. Hence $H_{\mathbb{C}}={\mathrm{\Psi }}_E(H\times H)$, and ${\mathrm{\Phi }}_E$ sends it exactly onto $H\times H$. Applying the same isomorphism to the real vector space $H$ identifies its abstract complexification $\mathbb{C}{\otimes }_{\mathbb{R}}H$ with $H\times H$; composing with ${\mathrm{\Psi }}_E$ gives an injective complex-linear map into $E_{\mathbb{C}}$, sending $z\otimes h$ to $z(1\otimes h)$. Thus the specified subspace is precisely the complexification induced by the inclusion of $H$. Bijection then identifies the removed unions and their complements when $T=E$.

The action axioms hold on each coordinate of the diagonal map. For each $g$, that map sends $H\times H$ to $(gH)\times (gH)$ and $T\times T$ to itself. Because $g$ permutes $\mathcal{A}$, it preserves the removed union and its complement. Its inverse is the diagonal map for $g^{-1}$.

To justify continuity, use the chosen linear coordinates $E\cong {\mathbb{R}}^n$ and the Euclidean norm in those coordinates. Each $g$ is represented by a linear map ${\mathbb{R}}^n\to {\mathbb{R}}^n$, so the Euclidean linear-map bound supplies $K_g\ge 0$ with $\mathrm{\parallel }gv{\mathrm{\parallel }}_2\le K_g\mathrm{\parallel }v{\mathrm{\parallel }}_2$. The bounded-linear-operator equivalences imply that $g$ is continuous; they apply to the Euclidean normed spaces here and also to the linear inverse $g^{-1}$. This includes $n=0$.

The subspace topology therefore makes $g{\mathrm{\mid }}_T:T\to T$ and its inverse continuous. Basic open sets of $T\times T$ are rectangles $U\times V$ with $U,V$ open in $T$. The inverse image of such a rectangle under the diagonal map is $(g{\mathrm{\mid }}_T{)}^{-1}(U)\times (g{\mathrm{\mid }}_T{)}^{-1}(V)$, which is open. Preimages commute with unions, so the diagonal map is continuous on every product-open set. The same argument applies to $g^{-1}$. Restricting domain and codomain to the invariant subspace $M(T,\mathcal{A})$ preserves continuity by the subspace-topology characteristic property, so these mutually inverse restricted maps are homeomorphisms.

The covering-space condition is an additional neighbourhood condition on this restricted action, not a consequence asserted by the definition.

Every Euclidean linear map has a unique matrix and satisfies ‖Lh‖₂≤ K‖h‖₂ for some K≥0

For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent