diff --git a/kaplansky_direct_finite_odd_v2.txt b/kaplansky_direct_finite_odd_v2.txt
index 4da6182..fc61016 100644
--- a/kaplansky_direct_finite_odd_v2.txt
+++ b/kaplansky_direct_finite_odd_v2.txt
@@ -81,7 +81,7 @@ infinite-dimensional vector space over K, with G itself as a basis,
 plus a multiplication rule coming from the group.
 
 Concrete example. When G=Z and K is any field, the
-group algebra K[Z]bbZ] is (isomorphic to) the ring of Laurent
+group algebra K[Z]Z] is (isomorphic to) the ring of Laurent
 polynomials K[x,x^{-1}] in one variable: send the integer~n to~x^n.
 That ring is commutative, so left and right behave the same there.
 Not every group algebra is commutative; for example, if G is
@@ -318,7 +318,7 @@ simple.
 The graph Gamma is simple and triangle-free. Let
 [
  T={T⊆[1200]:|T|le 400},
- t=|T|=calT|=sum_{j=0}^{400}binom{1200}{j}.
+ t=|T|=T|=sum_{j=0}^{400}binom{1200}{j}.
 ]
 There exist coefficients alpha_i(T) and beta_j(T) in F_p,
 indexed by TinT, such that the mtimes m matrix
@@ -409,7 +409,7 @@ For m=binom{1200}{600} we can compare factorials directly:
  =prod_{u=401}^{600}(u)/(u+200)
  le (tfrac 57)^{100}.
 ]
-Here the inequality is a crude but adequate estimate: for
+Here the ineuality is a crude but adequate estimate: for
 ule 500, the factor u/(u+200) is at most 500/700=5/7; for
 uge501, the factor is at most 600/800=3/4<1, so it contributes
 less than one. Multiplying all one hundred factors gives the displayed
@@ -436,7 +436,7 @@ t/m<1/100: the low-rank fit uses far fewer summands than there are
 vertices. The second, coming from the group H below, is
 d/w>1/96: the idempotent each vertex carries has a regular rank
 that is a fixed nonzero fraction of~|H|. Section~op:compression-section
-will combine the two ratios into a matrix with strict inequality
+will combine the two ratios into a matrix with strict ineuality
 wt<md between its shapes.
 
 ## The finite group H
@@ -509,7 +509,7 @@ group law in H gives the explicit form
 Indeed (c,0)(0,s)=(c,s) and (c,s)(-c,0)=(c+s·(-c),,s)=(c-s· c,s).
 
 
-The subgroups U and V satisfy Ucap V={1}, and Ucong Vcong P.
+The subgroups U and V satisfy Ucap V={1}, and U~= V~= P.
 
 Proof. 
 Suppose (0,0)=u(s)=v(s') for some s,s'in P. Comparing second
@@ -537,7 +537,7 @@ onto the other. This asymmetry is what makes the later triangular
 compression work.
 
 That completes the finite data. In later sections, we always write
-m=binom{1200}{600}, t=|T|,calT|, w=|H|, and
+m=binom{1200}{600}, t=|T|,T|, w=|H|, and
 d=dim_KeK[H] for the regular rank of the idempotent e we
 are about to build in Section~sec:modular.
 
@@ -547,7 +547,7 @@ are about to build in Section~sec:modular.
 This section is the technical heart of the paper. In the previous
 section we built three pieces of finite data: a prime p, a field
 K of order~p^4, and a finite group H=A_0rtimes P with two
-different subgroups U,Vcong P meeting trivially. Now we use all that
+different subgroups U,V~= P meeting trivially. Now we use all that
 data to build one particular idempotent ein K[H] with two properties:
  Large regular rank. The dimension d=dim_K(eK[H])
       of the image of left multiplication by e satisfies
@@ -581,12 +581,12 @@ A_0⊆ H.
 
 
 Let ell=(p-1)/2 and D=3(p-1). Work in the K-vector space K^P
-of functions Pto K, with standard basis mathbf e_xbfe_x (a 1 at
+of functions Pto K, with standard basis e_x (a 1 at
 x and 0 elsewhere). We want a linear subspace on which left shifts
 and right shifts both annihilate ell-fold.
 
 Translation calculus. For sin P let L_s:K^Pto K^P be the
-shift operator L_smathbf e_x=bfe_x=mathbf e_bfe_{x+s}. Define three operators
+shift operator L_se_x=e_x=e_{x+s}. Define three operators
 [
  X_j=L_{ε_j}-1_{K^P},  j=1,2,3,
 ]
@@ -606,26 +606,26 @@ generators of P.
 [Monomial basis]
 The p^3 vectors
 [
- X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0,bfe_0,  0le n_1,n_2,n_3< p,
+ X_1^{n_1}X_2^{n_2}X_3^{n_3}e_0,e_0,  0le n_1,n_2,n_3< p,
 ]
 form a K-basis of K^P.
 
 Proof. 
-The standard basis {mathbf e_x:xbfe_x:xin P} already has cardinality
-p^3=dim_KK^P. So it suffices to show every mathbf e_sbfe_s is in the
+The standard basis {e_x:xbfe_x:xin P} already has cardinality
+p^3=dim_KK^P. So it suffices to show every e_s is in the
 K-span of the displayed vectors.
 
 Fix s=(s_1,s_2,s_3)in P. Since 1+X_j=L_{ε_j}, we have
 (1+X_j)^{s_j}=L_{s_jε_j} and
 [
- mathbf e_s=L_sbfe_s=L_smathbf e_0bfe_0
- =prod_{j=1}^3 L_{s_jε_j}mathbf e_0bfe_0
- =prod_{j=1}^3 (1+X_j)^{s_j}mathbf e_0bfe_0
- =prod_{j=1}^3 (sum_{k=0}^{s_j}binom{s_j}{k}X_j^k)mathbf e_0.bfe_0.
+ e_s=L_sbfe_s=L_sbfe_0
+ =prod_{j=1}^3 L_{s_jε_j}e_0
+ =prod_{j=1}^3 (1+X_j)^{s_j}e_0
+ =prod_{j=1}^3 (sum_{k=0}^{s_j}binom{s_j}{k}X_j^k)e_0.e_0.
 ]
 The last product expands to a K-linear combination of the vectors
-X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0bfe_0 with 0le n_jle s_j<p.
-So mathbf e_sbfe_s lies in the span of the display. Since {mathbf e_sbfe_s}
+X_1^{n_1}X_2^{n_2}X_3^{n_3}e_0 with 0le n_jle s_j<p.
+So e_s lies in the span of the display. Since {e_s}
 spans, the display spans, and it has p^3 elements, so it is a basis.
 
 
@@ -639,7 +639,7 @@ truncated by X_j^p=0.
 Define B as the matrix with columns the basis vectors whose total
 degree is large enough:
 
- B := [X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0]_bfe_0]_{
+ B := [X_1^{n_1}X_2^{n_2}X_3^{n_3}e_0]_bfe_0]_{
  substack{0le n_1,n_2,n_3<p n_1+n_2+n_3ge D-ell+1}}.
 
 So B is a p^3times h matrix with K-linearly independent columns
@@ -664,7 +664,7 @@ ellth power (L_s-1)^ell has the property: applied to any monomial
 of total X-degree d, the result lies in the K-span of monomials
 of total X-degree at least d+ell.
 
-Now take a column X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0bfe_0 of B,
+Now take a column X_1^{n_1}X_2^{n_2}X_3^{n_3}e_0 of B,
 with d=n_1+n_2+n_3ge D-ell+1. Applying (L_s-1)^ell gives a
 combination of monomials of total degree at least D+1. But every
 monomial of total degree greater than D=3(p-1) has some exponent
@@ -672,8 +672,8 @@ at least p, and X_j^p=0 kills it. So each column is annihilated,
 proving the first identity.
 
 Second identity. Compute the transpose of L_s: its matrix entry
-(L_s)_{mathbf e_x,bfe_x,mathbf e_y}bfe_y (row x, column y) is 1 when
-x+s=y and 0 otherwise. Hence (L_s^{T}{mathbf e_y,bfe_y,mathbf e_x}bfe_x
+(L_s)_{e_x,e_x,e_y}e_y (row x, column y) is 1 when
+x+s=y and 0 otherwise. Hence (L_s^{T}{e_y,e_y,e_x}e_x
 is 1 when x-s=y, so L_s^{T}{-s}. Consequently
 (L_s-1)^{T}{-s}-1, and
 [
@@ -795,7 +795,7 @@ most (p^3-1)M^{p^2}, i.e. a fraction at most
  (p^3-1),M^{-p^2(p-1)}
  le (p^3-1)/M < 1/4 .
 ]
-The last inequality uses M=p^4-1>4p^3, which holds for pge5;
+The last ineuality uses M=p^4-1>4p^3, which holds for pge5;
 we have p>600.
 
 Singular N_lambda. The entries of D_lambda^{-1} range
@@ -923,17 +923,17 @@ Left multiplication by T_lambda on Mat_{p^3}(K) has image the
 matrices whose p^3 columns all lie in~im B, of dimension p^3h.
 Summing over the blocks:
 [
- d:=dim_K(eK[H])=p^3,h,|R|.scrR|.
+ d:=dim_K(eK[H])=p^3,h,|R|.R|.
 ]
 The admissible characters form more than half of A_0hat
 (Lemma~lem:good-count), and each P-orbit has size p^3, so
 [
- |R|>(|A_0|)/(2p^3).scrR|>(|A_0|)/(2p^3).
+ |R|>(|A_0|)/(2p^3).R|>(|A_0|)/(2p^3).
 ]
 Recall w=|H|=|A_0|,p^3. Therefore
 [
- frac dw=(p^3h|R|)/(|A_0|p^3)scrR|)/(|A_0|p^3)
- =(h|R|)/(|A_0|)scrR|)/(|A_0|)
+ frac dw=(p^3h|R|)/(|A_0|p^3)R|)/(|A_0|p^3)
+ =(h|R|)/(|A_0|)R|)/(|A_0|)
  >(h)/(2p^3)>frac1{96},
 ]
 using h/p^3>1/48 from Lemma~lem:h.