1\section{Proofs}2\label{app:proofs}34Throughout, $B$ is the finite set of bottleneck task classes, $\Omega(s)$ the partition across skills,5$\Qt = Q+\sum_s\CEUt(s)H_s$, and the composite $X_B(\{X_i\}) = (\sum_{i\in B}\nu_iX_i^{\rho_B})^{1/\rho_B}$ with6$\rho_B<0$ is concave, homogeneous of degree one and increasing, with every task essential7($X_B\to0$ as any $X_i\to0$).89\subsection*{Proof of Lemma~\ref{lem:ceu}}1011\emph{Step 1 (dual of the composite).} For task prices $p_i>0$ the unit cost of the composite is12$P_B(p)=[\sum_i\nu_i^{\sigma_B}p_i^{1-\sigma_B}]^{1/(1-\sigma_B)}$ and the cost-minimizing demands are13$X_i = \nu_i^{\sigma_B}(p_i/P_B)^{-\sigma_B}X_B$ (Shephard's lemma). If every task is produced with compute at price14$A$ per unit, $p_i = A\alpha_i$, so $P_B = A/\mathcal{B}$ with $\mathcal{B}$ as in \eqref{eq:XB}, and the compute15required per unit of composite is $\sum_i\alpha_iX_i/X_B = P_B/A = 1/\mathcal{B}$.1617\emph{Step 2 (pooling).} Write $\Qt_i = Q_i+\alpha_iL_i$ for the effective compute in task $i$. The planner's18problem $\max X_B$ subject to $\sum_iQ_i\le Q$, $\sum_{i\in\Omega(s)}L_i\le H_s$ is equivalent to19$\max_{\{\Qt_i\}}X_B(\{\Qt_i/\alpha_i\})$ subject to $\sum_i\Qt_i\le Q+\sum_i\alpha_iL_i$ for a given labor20allocation, and the inner maximum equals $\mathcal{B}(Q+\sum_i\alpha_iL_i)$ by Step 1 and homogeneity. The outer21maximum over labor assigns all of $H_s$ to $\argmax_{i\in\Omega(s)}\alpha_i$, giving $X_B=\mathcal{B}\Qt$.2223\emph{Step 3 (feasibility of the pooled allocation for $Q$ large).} The pooled allocation is feasible as an24allocation of compute and labor only if, in each task, the effective compute demanded exceeds the labor25assigned: $\Qt_i^*\ge\alpha_iL_i$. By Step 1, $\Qt_i^* = \alpha_iX_i^* = \alpha_i\nu_i^{\sigma_B}(\alpha_i\mathcal{B})^{-\sigma_B}\mathcal{B}\Qt$,26which is linear in $\Qt$, while $\alpha_iL_i\le\alpha_iH_{s(i)}$ is bounded. Hence there is a finite $\bar Q$27beyond which the constraint is slack in every task; for $Q\ge\bar Q$ the pooled allocation is feasible, hence28optimal, and compute is used in every task ($Q_i^* = \Qt_i^*-\alpha_iL_i>0$). For $Q<\bar Q$ some constraints29bind and the composite is strictly below $\mathcal{B}\Qt$ (Restrepo's transition).3031\emph{Step 4 (prices).} Competitive factor prices are the multipliers of the planner's problem. By the envelope32theorem $A = Z F_X\,\partial X_B/\partial Q = ZF_X\mathcal{B}$, and the multiplier on $H_s$ is33$ZF_X\mathcal{B}\max_{i\in\Omega(s)}\alpha_i = A\,\CEUt(s)$. \qed3435\subsection*{Proof of Proposition~\ref{prop:laborshare}}3637By Lemma~\ref{lem:ceu}, $\sum_sw_sH_s = A\sum_s\CEUt(s)H_s$ and $A\Qt = ZF_X\mathcal{B}\Qt = ZF_XX_B$.38Since $F$ is homogeneous of degree one and increasing, Euler's theorem gives $Y = ZF_XX_B+ZF_SS\ge ZF_XX_B = A\Qt$.39Hence $\sum_sw_sH_s/Y\le A\sum_s\CEUt(s)H_s/(A\Qt) = \sum_s\CEUt(s)H_s/\Qt$, which goes to zero as $Q\to\infty$40because the numerator is finite under Premise~1. The argument uses only $Y\ge ZF_XX_B$, which holds for any41constant-returns $F$ increasing in all arguments; it does not use the form of $F$, the number of complements, or42how $Q$ is produced. \qed4344\subsection*{Proof of Proposition~\ref{prop:regimes}}4546Write $x\equiv\mathcal{B}\Qt$. (a) $\rho\in(0,1)$: $F = x[\nu+(1-\nu)(S/x)^{\rho}]^{1/\rho}$, so47$F/x\to\nu^{1/\rho}$ and $F_X = \nu x^{\rho-1}F^{1-\rho}\to\nu\cdot\nu^{(1-\rho)/\rho} = \nu^{1/\rho}$;48hence $A\to Z\nu^{1/\rho}\mathcal{B}$ and $s_Q = AQ/Y\to1$ because $Y/Q\to Z\nu^{1/\rho}\mathcal{B}$ and49$\Qt/Q\to1$. (b) $\rho=0$: $F = x^{\nu}S^{1-\nu}$, $F_X = \nu F/x\to0$ as $x\to\infty$; the shares are50$\nu$ and $1-\nu$ identically. (c) $\rho<0$: $x^{\rho}\to0$ so $F\to(1-\nu)^{1/\rho}S$ and51$F_X = \nu x^{\rho-1}F^{1-\rho}\to0$ at rate $x^{\rho-1}$; the share of $S$, $(1-\nu)S^{\rho}/[\nu x^{\rho}+(1-\nu)S^{\rho}]$,52converges to one. Wages are $A\,\CEUt(s)$ in all cases. For the last claim, the pre-AGI wage bill is positive53and independent of $Q$ while the post-AGI wage bill is $A\sum_s\CEUt(s)H_s\to0$ in (b) and (c). \qed5455\subsection*{Proof of Proposition~\ref{prop:bgp}}5657Guess a path on which $S_t$, $K^Q_t$, $Y_t$ and capitalists' expenditure grow at a common rate $g$ and58$\Qt_t/S_t$ is constant (labor's compute-equivalent is negligible asymptotically, so $\Qt_t/K^Q_t\to1$).59Constant returns of $F$ give a constant $Y/S$, hence constant marginal products $R_S = ZF_S$ and $A = ZF_X\mathcal{B}$.60The two capital Euler equations without adjustment costs read $1+r = R_S+1-\delta_S$ and61$1+r = (A + (1-\delta_Q)\bar p)/\bar p$, i.e.\ $A = (r+\delta_Q)\bar p$, and the capitalists' Euler equation62gives $1+r = \beta^{-1}(1+g)^{\gamma}$. These three equations in $(r,g,\Qt/S)$ have a solution because63$R_S(\Qt/S)$ is increasing in $\Qt/S$ (complementarity) and $A(\Qt/S)$ is decreasing, so the ratio is pinned by64$A/R_S = (r+\delta_Q)\bar p/(r+\delta_S)$, and then $g$ follows from the Euler equation. The investment rates are65constant and the goods market clears by Walras' law. The share $s_Q$ is the CES share formula evaluated at the66constant ratio; it is strictly between zero and one. The labor share converges to zero by67Proposition~\ref{prop:laborshare}. \qed6869\subsection*{Proof of Proposition~\ref{prop:moore}}7071(a) With $\rho\in(0,1)$ and $S$ growing at most at the rate of accumulation out of a finite saving share,72$F/\Qt\to Z\nu^{1/\rho}\mathcal{B}$ as $\Qt/S\to\infty$, so $A$ has a positive limit. The compute Euler equation73$A_{t+1} = (1+r_{t+1})p^Q_t-(1-\delta_Q)p^Q_{t+1}$ with $p^Q_t\to0$ then requires $r_{t+1}\to\infty$, and the74Euler equation $1+r_{t+1} = \beta^{-1}(1+g_{t+1})^{\gamma}$ forces the growth rate of capitalists' expenditure, and75hence of output, to diverge. A balanced path with finite $g$ is impossible. (b) With $\rho\le0$ and76$\Qt/S\to\infty$ (compute is free at the margin), $F\to(1-\nu)^{1/\rho}S$ for $\rho<0$ and $F = x^{\nu}S^{1-\nu}$ for77$\rho=0$; in both cases $F_X\to0$ so $A\to0$ and $w_s = A\,\CEUt(s)\to0$. The value of compute capital is78$p^Q_tK^Q_t = A_tK^Q_t/[(1+r)(p^Q_{t-1}/p^Q_t)-(1-\delta_Q)]$ and $A_tK^Q_t/Y_t$ is the compute share, which goes to79zero for $\rho<0$ and is $\nu$ for $\rho=0$; in the Cobb--Douglas case the market value of compute relative to80output is $\nu/[(1+r)e^{g_{pQ}}-(1-\delta_Q)]$, bounded, while in the complements case it goes to zero. \qed8182\subsection*{Proof of Proposition~\ref{prop:energy}}8384Energy feasibility gives $\varepsilon_tQ_t\le\bar E$, so $Q_t\le\bar E/\bar\varepsilon$ and $\Qt_t$ is bounded.85Since $F_X>0$ at any bounded $\Qt$, $A_t = ZF_X\mathcal{B}$ is bounded below. The labor share equals86$s_{Q,t}\sum_s\CEUt(s)H_s/\Qt_t$ by Lemma~\ref{lem:ceu} (with $s_{Q,t} = A_t\Qt_t/Y_t$ the share of the task87composite), and $\Qt_t\le\bar E/\bar\varepsilon+\sum_s\CEUt(s)H_s$ gives the bound. \qed8889\subsection*{Proof of Proposition~\ref{prop:wealth}}9091On a balanced path without adjustment costs the ex-dividend value of capital type $k$ is92$V_k = p_kK'_k = p_kK_k(1+g)$ and the Euler equation gives $R_k = (1+r)p_{k,-1}-(1-\delta_k)p_k =93p_k[(1+r)(1+\pi_k)-(1-\delta_k)]$ with $p_{k,-1}/p_k = 1+\pi_k$. Hence $V_k/Y = s_k(1+g)/[(1+r)(1+\pi_k)-(1-\delta_k)]$94with $s_k = R_kK_k/Y$. Land: $P_T = \sum_{j\ge1}R_T(1+g_T)^j/(1+r)^j = R_T(1+g_T)/(r-g_T)$. Summing gives95\eqref{eq:WY}. The comparative static follows because the coefficient on $s_k$ is $1/[(1+r)(1+\pi_k)-(1-\delta_k)]$,96decreasing in $\delta_k$ and in $\pi_k$. The price--earnings ratio is $V_k/(R_kK_k)$. \qed9798\subsection*{Proof of Proposition~\ref{prop:discount}}99100With a constant price index the capitalists' Euler equation gives $1+r = \beta^{-1}(1+g)^{\gamma}$. For a cash101flow growing at $g$, $P/D = (1+g)/(r-g) = (1+g)/[\beta^{-1}(1+g)^{\gamma}-(1+g)]$. Let102$h(g) = \log(1+g)-\log[\beta^{-1}(1+g)^{\gamma}-(1+g)]$. Differentiating with respect to $\log(1+g)$,103\[104\frac{\dd h}{\dd\log(1+g)} = 1-\frac{\gamma\beta^{-1}(1+g)^{\gamma}-(1+g)}{\beta^{-1}(1+g)^{\gamma}-(1+g)}105= \frac{(1-\gamma)\,\beta^{-1}(1+g)^{\gamma}}{\beta^{-1}(1+g)^{\gamma}-(1+g)},106\]107whose denominator is $(1+g)(r-g)/(1+g)\cdot(1+g)>0$ whenever $r>g$ (the transversality condition). The sign is108therefore that of $1-\gamma$. \qed109110\subsection*{Proof of Proposition~\ref{prop:distribution}}111112Workers' expenditure is $\sum_sw_sH_s+\tau_t$; with $\tau_t=0$ and $Y_t\to\infty$ the ratio to $Y_t$ is the113labor share, which goes to zero by Proposition~\ref{prop:laborshare}; the level converges to114$A\sum_s\CEUt(s)H_s$ by Lemma~\ref{lem:ceu}. With the fund, $\tau_t = \theta_F\mu_WD_t$ where $D_t$ is the115aggregate net payout; on the balanced path $D_t/Y_t = \bar d$, which gives the bound. \qed116117\subsection*{Proof of Proposition~\ref{prop:tax}}118119With the tax, the after-tax rental received by the owner of a unit of capacity is $(1-\tau_Q)(A-\varepsilon P_E)$120and the Euler equation on the balanced path is $(1-\tau_Q)(A-\varepsilon P_E) = (r+\delta_Q)\bar p$. Solving for121$A$ and using $w_s = A\,\CEUt(s)$ gives the expression. A tax on land rent at rate $\tau_T$ rebated lump-sum122changes no first-order condition of any producer or investor other than the land-price recursion, in which the123after-tax rent replaces the rent; the stock of land is fixed so allocations are unchanged. \qed124