[proofplan] The proof combines a minimality argument with Euler-type edge bounds on surfaces. Let $k = \chi(G)$ and pass to a chromatic-critical subgraph $G'$ — a subgraph of chromatic number $k$ that is minimal under vertex deletion. This forces $\delta(G') \geq k - 1$ and $|G'| \geq k$. The surface Euler characteristic gives the edge bound $e(G') \leq 3(|G'| - \mathcal{E})$, which by double-counting yields an average-degree bound of $6 - 6\mathcal{E}/|G'|$. Combining $\delta(G') \leq \operatorname{avg deg}(G')$ with $|G'| \geq k$ and $\mathcal{E} \leq 0$ gives the quadratic inequality $k^2 - 7k + 6\mathcal{E} \leq 0$; solving yields $k \leq H(\mathcal{E})$. [/proofplan]

[step:Reduce to a chromatic-critical subgraph with minimum degree at least $k - 1$]Let $k := \chi(G)$. Among all subgraphs of $G$ with chromatic number equal to $k$, choose one with the fewest vertices; call it $G'$. Then $\chi(G') = k$ and for every $v \in V(G')$, $\chi(G' - v) < k$. **Claim:** $\delta(G') \geq k - 1$. *Proof of claim.* Suppose for contradiction that some vertex $v \in V(G')$ satisfies $\deg_{G'}(v) \leq k - 2$. By minimality, $\chi(G' - v) \leq k - 1$, so $G' - v$ admits a proper colouring with $k - 1$ colours. Since $v$ has at most $k - 2$ neighbours in $G'$, its neighbours use at most $k - 2$ of the $k - 1$ available colours, leaving at least one colour free to assign to $v$. Extending the colouring to $v$ yields a proper $(k-1)$-colouring of $G'$, contradicting $\chi(G') = k$. Hence $\delta(G') \geq k - 1$. **Also: $|G'| \geq k$.** Any graph with chromatic number $k$ has at least $k$ vertices, since a proper colouring with fewer than $k$ colours would exist directly on fewer than $k$ vertices. (More precisely: $\chi(G') \leq |G'|$ always, so $k = \chi(G') \leq |G'|$.) $G'$ remains drawn on the same surface as $G$ (subgraphs of drawings are drawings).[/step]

[guided]The reduction to a chromatic-critical subgraph is standard: a counterexample to "every graph on the surface has chromatic number at most $H(\mathcal{E})$" with the fewest vertices must be **critical** — removing any vertex strictly reduces the chromatic number. Critical graphs have structural constraints (here, $\delta \geq k - 1$) that the original graph may lack, and these constraints are what we exploit. **Why $\delta(G') \geq k - 1$?** Imagine a vertex $v$ of degree $\leq k - 2$. We would be able to colour $G' - v$ with $k - 1$ colours (by criticality), then extend to $v$: $v$ has at most $k - 2$ neighbours, using at most $k - 2$ of the $k - 1$ colours, so at least one free colour remains. This gives $G'$ a $(k-1)$-colouring, contradicting $\chi(G') = k$. So every vertex must have degree at least $k - 1$. **Why $|G'| \geq k$?** If $|G'| < k$, a proper colouring using one colour per vertex yields $\chi(G') \leq |G'| < k$, contradicting $\chi(G') = k$. It remains to convert these bounds into a numerical inequality for $k$, which we do via the surface Euler inequality.[/guided]

[step:Apply the surface edge bound $e(G') \leq 3(|G'| - \mathcal{E})$]For a connected simple graph drawn on a compact orientable surface of Euler characteristic $\mathcal{E}$ with at least three vertices, the Euler formula $|V| - |E| + |F| = \mathcal{E}$ combined with the face-degree inequality $2|E| = \sum_F \deg F \geq 3|F|$ yields \begin{align*} e(G') \leq 3(|G'| - \mathcal{E}). \end{align*} We verify the hypotheses: $|G'| \geq k$ (from the previous step) and $k \geq 3$ (since $\mathcal{E} \leq 0$, the sphere case $\mathcal{E} = 2$ is excluded; also $|G| \geq 3$ is hypothesized and passes through criticality when $k \geq 3$, which we establish separately below).[/step]

[guided]**Why $k \geq 3$?** We need this to guarantee $|G'| \geq 3$, which is the hypothesis of the surface edge bound. If $k \leq 2$, then $\chi(G) \leq 2$, and we check directly that $H(\mathcal{E}) \geq 2$ for all $\mathcal{E} \leq 0$. Indeed, at $\mathcal{E} = 0$, $H(0) = \lfloor (7 + 7)/2 \rfloor = 7$; as $\mathcal{E}$ decreases, $H(\mathcal{E})$ increases. So $\chi(G) \leq 2 \leq H(\mathcal{E})$ holds directly, and the non-degenerate regime is $k \geq 3$. **Why the face-degree inequality $\sum \deg F \geq 3|F|$?** Each face of a simple graph drawing has boundary a closed walk of length $\geq 3$: faces of length $1$ are loops (not present in simple graphs), and faces of length $2$ are double edges (not present). Hence every face contributes at least $3$ to $\sum_F \deg F$. Summing, $2|E| = \sum_F \deg F \geq 3|F|$, giving $|F| \leq \tfrac{2}{3}|E|$. **Why $|V| - |E| + |F| = \mathcal{E}$?** This is the defining Euler-characteristic formula for a cellular embedding on a compact orientable surface. The hypothesis "drawn on a compact orientable surface of Euler characteristic $\mathcal{E}$" is precisely what lets us invoke this. If the embedding is not cellular, $|V| - |E| + |F| \geq \mathcal{E}$ still holds (replacing equality by inequality only strengthens what follows), since any embedding can be refined to a cellular one by adding cells without increasing $|V| - |E| + |F|$. Substituting $|F| \leq \tfrac{2}{3}|E|$ into $|V| - |E| + |F| \geq \mathcal{E}$ gives $|V| - |E| + \tfrac{2}{3}|E| \geq \mathcal{E}$, so $|V| - \tfrac{1}{3}|E| \geq \mathcal{E}$, hence $|E| \leq 3(|V| - \mathcal{E})$. This is the edge bound we needed.[/guided]

[step:Bound the average degree of $G'$ via double-counting]By the handshake identity, \begin{align*} \sum_{v \in V(G')} \deg_{G'}(v) = 2\,e(G') \leq 6(|G'| - \mathcal{E}). \end{align*} Dividing by $|G'|$, \begin{align*} \operatorname{avgdeg}(G') = \frac{1}{|G'|}\sum_{v \in V(G')} \deg_{G'}(v) \leq \frac{6(|G'| - \mathcal{E})}{|G'|} = 6 - \frac{6\mathcal{E}}{|G'|}. \end{align*} Since the minimum degree is at most the average degree, \begin{align*} \delta(G') \leq 6 - \frac{6\mathcal{E}}{|G'|}. \end{align*} Using $|G'| \geq k$ and $\mathcal{E} \leq 0$ (so that $-\mathcal{E} \geq 0$ and thus $-6\mathcal{E}/|G'|$ is maximised by the smallest valid $|G'|$, namely $|G'| = k$): \begin{align*} \delta(G') \leq 6 - \frac{6\mathcal{E}}{k}. \end{align*}[/step]

[guided]The direction of the inequality matters because $\mathcal{E} \leq 0$, so $-\mathcal{E} \geq 0$ and $-6\mathcal{E}/|G'|$ is a non-negative quantity. Since we want an upper bound on $\delta(G')$, we use the **largest** possible value of $-6\mathcal{E}/|G'|$, which occurs at the **smallest** possible $|G'|$. The bound $|G'| \geq k$ tells us the smallest possible $|G'|$ is $k$, so we substitute $|G'| = k$: \begin{align*} 6 - \frac{6\mathcal{E}}{|G'|} \leq 6 - \frac{6\mathcal{E}}{k}. \end{align*} If $\mathcal{E} > 0$ (the sphere case), the direction of the inequality would flip, and we would need the **largest** $|G'|$, not the smallest. That is why the hypothesis $\mathcal{E} \leq 0$ is essential: the theorem is about surfaces other than the sphere. Combining with $\delta(G') \geq k - 1$ from the critical-subgraph step: \begin{align*} k - 1 \leq \delta(G') \leq 6 - \frac{6\mathcal{E}}{k}. \end{align*}[/guided]

[step:Solve the resulting quadratic inequality for $k$]Combining the two-sided estimate $k - 1 \leq 6 - 6\mathcal{E}/k$ and multiplying through by $k > 0$: \begin{align*} k(k-1) &\leq 6k - 6\mathcal{E} \\ k^2 - k - 6k + 6\mathcal{E} &\leq 0 \\ k^2 - 7k + 6\mathcal{E} &\leq 0. \end{align*} Treating this as a quadratic in $k$ and completing the square: \begin{align*} \left(k - \tfrac{7}{2}\right)^2 = k^2 - 7k + \tfrac{49}{4} \leq \tfrac{49}{4} - 6\mathcal{E}. \end{align*} Taking square roots (valid since both sides are non-negative: the right-hand side is $\tfrac{49}{4} - 6\mathcal{E} \geq \tfrac{49}{4} > 0$ because $\mathcal{E} \leq 0$), \begin{align*} \left|k - \tfrac{7}{2}\right| \leq \tfrac{1}{2}\sqrt{49 - 24\mathcal{E}}, \end{align*} hence \begin{align*} k \leq \tfrac{7}{2} + \tfrac{1}{2}\sqrt{49 - 24\mathcal{E}} = \frac{7 + \sqrt{49 - 24\mathcal{E}}}{2}. \end{align*} Since $k = \chi(G)$ is an integer, $k \leq \left\lfloor \tfrac{7 + \sqrt{49 - 24\mathcal{E}}}{2} \right\rfloor = H(\mathcal{E})$.[/step]

[guided]The quadratic $k^2 - 7k + 6\mathcal{E} = 0$ has roots \begin{align*} k = \frac{7 \pm \sqrt{49 - 24\mathcal{E}}}{2}. \end{align*} The inequality $k^2 - 7k + 6\mathcal{E} \leq 0$ is satisfied between the two roots, so $k$ lies in the interval \begin{align*} \left[ \frac{7 - \sqrt{49 - 24\mathcal{E}}}{2}, \frac{7 + \sqrt{49 - 24\mathcal{E}}}{2} \right]. \end{align*} The lower endpoint is negative or small (for $\mathcal{E} = 0$, it is $0$; for $\mathcal{E} < 0$, it is negative), so it imposes no constraint on the positive integer $k$. The upper endpoint is the operative bound. **Floor function.** $k$ is an integer and $k \leq \tfrac{7 + \sqrt{49 - 24\mathcal{E}}}{2}$, so $k \leq \lfloor \tfrac{7 + \sqrt{49 - 24\mathcal{E}}}{2} \rfloor = H(\mathcal{E})$. **Numerical check: torus ($\mathcal{E} = 0$).** $H(0) = \lfloor (7 + 7)/2 \rfloor = 7$: every graph on the torus is $7$-colourable, which is a classical result (and tight for $K_7$ on the torus). **Numerical check: Klein bottle ($\mathcal{E} = 0$ for non-orientable; but the theorem is about orientable).** The orientable version gives $7$ as the chromatic bound; this is tight. **The direction of the bound on $\delta$ versus avg degree.** We used $\delta \leq$ avg degree, which is always true. The other direction is not true in general. Because we derived $\delta \geq k - 1$ from critical-subgraph analysis, the chain $k - 1 \leq \delta \leq$ avg degree $\leq 6 - 6\mathcal{E}/k$ produces the desired inequality.[/guided]

[step:Conclude $\chi(G) \leq H(\mathcal{E})$] Since $\chi(G) = k \leq H(\mathcal{E})$, the theorem is proved for all connected graphs $G$ with $|G| \geq 3$ drawn on a compact orientable surface of Euler characteristic $\mathcal{E} \leq 0$. [/step]