diff --git a/disctospec.tex b/disctospec.tex
index df477b7336d77200f0374eba2acbc430b994e362..46662ce3fd7f4ec564818be6b2ea92e2e78e17b6 100644
--- a/disctospec.tex
+++ b/disctospec.tex
@@ -105,6 +105,7 @@ By convexity,~\eqref{eq:reduce} is maximized when $x'[j] \in \{0, 1\}^n$ for all
 
 By a straightforward generalization of the Expander Mixing Lemma, it follows that if a Cayley hypergraph has spectral expansion $\lambda$, then it also has discrepancy $\lambda$.  We attempt to prove the converse, stated below.
 
+\snote{This is misleading!}
 \begin{lem}\label{lem:disctospec}
 Let $\Gamma = F_p^n$. For $t \leq p$ and every $\lambda$ there exists an $\epsilon$ such that if a Cayley $t$-uniform hypergraph $H = \cay^{(t)}(\Gamma, q, S')$ has discrepancy $\epsilon$ with respect to $K = \cay^{(t)}(\Gamma, q, S)$, then it also has spectral expansion $\lambda$ with respect to $K = \cay^{(t)}(\Gamma, q, S)$.
 \end{lem}
@@ -203,12 +204,13 @@ We prove the contrapositive.  Assume that $\cay^{(t)}(\Gamma, q, S')$ does not h
  \begin{equation}
 (A_H-A_K)(x_1, \ldots, x_t) = \frac{|\Gamma|^2}{t!} \sum_{\sigma \in S_t} \mathbb{E}_{u, g}[x_1(ug^{\sigma(1)-1})x_2(ug^{\sigma(2)-1})\cdots x_t(ug^{\sigma(t)-1})1_{H, K}(g)]. \label{eq:unravel}
 \end{equation}
-  By the Lemma~\ref{lem:genvn},~\eqref{eq:unravel} is bounded above by $\|1_{H, K}\|_{U^t}$.  By the inverse Gowers theorem, there exists a polynomial $P$ of degree $t-1$ such that $\mathbb{E}_{h}[e(P(g)) 1_{H, K}] \geq \epsilon$ for some $\epsilon$.  By Lemma~\ref{lem:decomp}, there exist $P_0, \ldots, P_{t-1}$ such that \[e(P_0(x))e(P_1(x+y))\cdots e(P_{t-1}(x+(t-1)y)) = e(P(y)),\] and therefore \[\mathbb{E}_{u, g}[e(P_0(u))e(P_1(ug))\cdots e(P_{t-1}(ug^{t-1})) 1_{H, K}] = \mathbb{E}_{h}[e(P(g)) 1_{H, K}] \geq \epsilon,\]
+  By the Lemma~\ref{lem:genvn},~\eqref{eq:unravel} is bounded above by $\Gamma|^2\|1_{H, K}\|_{U^t}$.  By the inverse Gowers theorem, there exists a polynomial $P$ of degree $t-1$ such that $\mathbb{E}_{h}[e(P(g)) 1_{H, K}] \geq \epsilon$ for some $\epsilon$, that depends on $\lambda/|\Gamma|^2$. \snote{This is really bad}
+  By Lemma~\ref{lem:decomp}, there exist $P_0, \ldots, P_{t-1}$ such that \[e(P_0(x))e(P_1(x+y))\cdots e(P_{t-1}(x+(t-1)y)) = e(P(y)),\] and therefore \[\mathbb{E}_{u, g}[e(P_0(u))e(P_1(ug))\cdots e(P_{t-1}(ug^{t-1})) 1_{H, K}] = \mathbb{E}_{h}[e(P(g)) 1_{H, K}] \geq \epsilon,\]
 and
 \[
 (A_H-A_K)(e(P_0), \ldots, e(P_{t-1})) \geq \frac{\epsilon|\Gamma|^2}{t!}.
 \]
-Because the coordinates of $e(P_i)$ are bounded above in absolute value by $1$ for all $i$, by Lemma~\ref{lem:disctoinfty}, this implies that $H$ has discrepancy at least $\epsilon|\Gamma|^2/(t!\pi^t)$ with respect to $K$.
+Because the coordinates of $e(P_i)$ are bounded above in absolute value by $1$ for all $i$, by Lemma~\ref{lem:disctoinfty}, this implies that $H$ has discrepancy at least $\epsilon|\Gamma|/(t!\pi^t)$ with respect to $K$.
 \end{proof}
 
 
