Fact 2.4. Every definable infinite set S in ${\mathcal {K}}$ is locally almost strongly internal to $K, \ \textbf {k}, \ \Gamma $ or $K/{\mathcal {O}}$ .

Fact 2.6 [Reference Halevi, Hasson and Peterzil13, Fact 4.25, Proposition 4.35].

Let G be an infinite A-definable group in ${\mathcal {K}}$ locally almost strongly internal to an unstable distinguished sort D. Then there is an A-definable finite normal abelian subgroup $H\trianglelefteq G$ such that $G/H$ is a D-group. Moreover,

1. (1) The almost D-rank and the D-rank of $G/H$ are equal (and equal to the almost D-rank of G).

2. (2) H is invariant under any definable automorphism of G and is contained in any definable finite index subgroup of G.

Proof. By dp-minimality and sub-additivity of dp-rank $\mathrm {dp\text {-}rk}(a/A)\leq \dim _{\operatorname {\mathrm {acl}}}(a/A)$ , proving $(2)\Rightarrow (1)$ . For the other direction, assume $(1)$ .

Let $a^{\prime }\subseteq a$ be $\operatorname {\mathrm {acl}}$ -independent over A of maximal length d – namely, $d=\dim _{\mathrm {ind}}(a/A)$ . Since $a^{\prime }$ is $\operatorname {\mathrm {acl}}$ -independent over $A, \ \dim _{\operatorname {\mathrm {acl}}}(a^{\prime }/A)=d$ , which by assumption equals $\mathrm {dp\text {-}rk}(a^{\prime }/A)$ . Thus, $\mathrm {dp\text {-}rk}(a/A)\geq \mathrm {dp\text {-}rk}(a^{\prime }/A)=d=\dim _{\mathrm {ind}}(a/A)$ , and equality of $\mathrm {dp\text {-}rk}$ and $\dim _{\mathrm {ind}}$ follows.

Fact 2.14 [Reference Halevi, Hasson and Peterzil13, Proposition 5.8].

Let D be an unstable distinguished sort and let G be a D-group.

1. (1) Assume that $X\subseteq G$ is a D-set over A. Then for every A-generic $a,b\in X$ , the set $\nu _X(a)a^{-1}$ is a (type-definable) subgroup of G and $\nu _X(a)a^{-1}=\nu _X(b)b^{-1}=a^{-1}\nu _X(a)$ . We denote this group $\nu _X$ .

2. (2) If $X,Y\subseteq G$ are D-sets over A, then $\nu _X=\nu _Y$ , and we can call it $\nu _D(G)$ , the infinitesimal type-definable subgroup of G with respect to D.

3. (3) For every $g\in G({\mathcal {K}})$ , we have $g\nu _D(G) g^{-1}=\nu _D(G)$ . In fact, $\nu _D$ is invariant under any ${\mathcal {M}}$ -definable automorphism of G.

Proof. Let $H\leq G$ be any subgroup, as in the statement.

(1) Let $X_G\subseteq G$ be a D-set in G and $X_H\subseteq H$ a D-set in H, all definable over a parameter set A. Let $(g,h)\in X_G\times X_H$ be generic over A, so $\nu _D(G)=g^{-1}\nu _{X_G}(g)$ and $\nu _D(H)=h^{-1}\nu _{X_H}(h)$ .

Let V be a generic vicinity of g and U a generic vicinity of h. By [Reference Halevi, Hasson and Peterzil13, Lemma 4.26], $U\cap hg^{-1} V$ is a generic vicinity of h, and hence,

$$\begin{align*}\nu_D(H)\vdash h^{-1}(U\cap hg^{-1}V)=h^{-1}U\cap g^{-1}V\subseteq g^{-1}V.\end{align*}$$

(2) Assume that H and G have the same D-rank; hence, any D-set in H is automatically a D-set in G. It now follows by definition that $\nu _D(H)=\nu _D(G)$ .

If H has finite index in G then it is easy to see that they have the same D-rank.

Proof. Let $Z^{\prime }$ be any D-set, definable over some parameter set $B^{\prime }$ . Find an element $g^{\prime }\equiv _A g$ such that $\mathrm {dp\text {-}rk}(g^{\prime }/AB^{\prime })=\mathrm {dp\text {-}rk}(g/A)$ . Applying an automorphism over A, we can move $g^{\prime }$ to g and $B^{\prime }$ to some B. The image, Z, of $Z^{\prime }$ under this automorphism, is definable over B and $\mathrm {dp\text {-}rk}(g/AB)=\mathrm {dp\text {-}rk}(g/A)$ . Renaming, we assume from now on, that $A=AB$ .

Fix an A-generic $h\in Z$ with $\mathrm {dp\text {-}rk}(g,h/A)=\mathrm {dp\text {-}rk}(X)+\mathrm {dp\text {-}rk}(Z)$ . Thus, as $\nu _D=h^{-1}\nu _Z(h)$ , we have to show that $\mathrm {dp\text {-}rk}(X\cap gh^{-1}\nu _Z(h))=\mathrm {dp\text {-}rk}(X)$ .

Let $Y\subseteq Z$ be some B-generic vicinity of h (i.e., $Y\in \nu _Z(h)$ ), for some B; so it will suffice to prove that $\mathrm {dp\text {-}rk}(X\cap gh^{-1}Y)=\mathrm {dp\text {-}rk}(X)$ .

By [Reference Halevi, Hasson and Peterzil13, Lemma 4.13], there exists $C\supseteq A$ and a C-generic vicinity $Y^{\prime }\subseteq Y$ of h such that $\mathrm {dp\text {-}rk}(g,h/A)=\mathrm {dp\text {-}rk}(g,h/C)$ . So $(g,h)$ is C-generic in $X\times Y^{\prime }$ . It will be sufficient to prove that $\mathrm {dp\text {-}rk}(X\cap gh^{-1}Y^{\prime })=\mathrm {dp\text {-}rk}(X)$ ; this is exactly [Reference Halevi, Hasson and Peterzil13, Lemma 4.26].

Proof. For (1) and (2), we first note that for any (almost) D-critical set $X\subseteq G$ , there exists an (almost) D-critical $Y\subseteq f(X)$ (with respect to $G/H$ ), with $\mathrm {dp\text {-}rk}(Y)=\mathrm {dp\text {-}rk}(X)$ . Indeed, if D is an SW-uniformity, then this is [Reference Halevi, Hasson and Peterzil13, Lemma 2.9], and if $D=\textbf {k}$ in the V-minimal case, then it is [Reference Halevi, Hasson and Peterzil13, Lemma 4.3]. This implies (1) and (2) for D other than $K/{\mathcal {O}}$ in the p-adically closed case. For (1) in that latter case, use [Reference Halevi, Hasson and Peterzil13, Lemma 3.9].

We now assume that D is not $K/{\mathcal {O}}$ in the p-adically closed case.

(3) If G is a D-group, then $G/H$ is also locally strongly internal to D by (2). Combined with (the proof of) [Reference Halevi, Hasson and Peterzil13, Fact 4.25], it follows that $G/H$ is also a D-group.

To show that $f(\nu _{D}(G))=\nu _{D}(G/H)$ , let $X_0\subseteq G$ be a D-set. By the above, we may find a D-critical subset $Y_0\subseteq f(X_0)$ . By [Reference Halevi, Hasson and Peterzil13, Remark 4.18], there exists a D-set $Y\subseteq Y_0\subseteq G/H$ . Setting $X=f^{-1}(Y)\subseteq X_0$ , and since $X_0$ is a D-set, so is $X_0$ . We are now in the situation where X and $Y=f(X)$ are both D-sets, with respect to G and $G/H$ , respectively. Assume everything is defined over some parameters set A.

Let $a\in X$ be an A-generic in X, so $f(a)$ is an A-generic in Y. It suffices to prove that $f(\nu _X(a))=\nu _X(f(a))$ .

For this, first note that if $U\subseteq X$ is a B-generic vicinity of a, for some $B\supseteq A$ , then $f(U)$ is a B-generic vicinity of $f(a)$ since $f(a)\in \operatorname {\mathrm {dcl}}(Aa)$ and $\mathrm {dp\text {-}rk}(U)=\mathrm {dp\text {-}rk}(f(U))$ as f is finite-to-one.

To show the other direction, let V be a B-generic vicinity of $f(a)$ for some $B\supseteq A$ . Then $f^{-1}(V)$ is a B-generic vicinity of a since $a\in \operatorname {\mathrm {acl}}(Af(a))$ and $f(f^{-1}(V))=V$ because f is surjective.

(4) Follows directly from (1) and (2),

Fact 2.19. Let G be a group definable in a sufficiently saturated NIP structure and $\{H_i:i\in T\}$ a definable family of finite index subgroups of G. Then $\bigcap _{i\in T} H_i$ is a definable subgroup of finite index.

Proof. By Baldwin-Saxl, there is a finite bound on the index of finite intersections of the $H_i$ .

Proof. By Fact 2.19, $C_G(X)$ has finite index in G. Applying this fact again to the intersection of the family $\{\lambda _t(C_G(X)):t\in T\}$ gives the desired conclusion.

Proof. (1) Since $N/H$ is abelian, for every $a,b\in N, \ ab=bah$ for some $h\in H$ . Because H is normal, for all $g\in G$ and $h\in H$ , there is $h^{\prime }\in H$ such that $hg=gh^{\prime }$ . It follows that $a^2b^2=(ab)^2h_1$ , for $h_1\in H$ , and by induction, $a^kb^k=(ab)^k h_0$ , for some $h_0\in H$ . Thus, $N^kH$ is a subgroup, clearly normal in N.

The order of every $g\in N/N^kH$ is at most k; thus, $N/N^kH$ has bounded exponent. The group $N/N^kH$ is clearly also definable in ${\mathcal {K}}$ , and by [Reference Halevi, Hasson and Peterzil13, Theorem 7.4, Theorem 7.7 and Theorem 7.11], a definable group of bounded exponent must be finite. Thus, $N/N^kH$ must be finite.

(2) Assume now that $k=|H|$ and H is central. Since $G/H$ is abelian, for every $g, x\in N$ , we have $g^{-1}xg=x h$ for some $h\in H$ , and hence, since H is central, $g^{-1}x^k g=(xh)^k=x^kh^k=x^k$ . Thus, $N^k\subseteq Z(N)$ . It follows that $N^kH\subseteq Z(N)$ , so by (1), $Z(N)$ has finite index in N.

Proof. For simplicity, let us call a set invariant under all the $\lambda _t \Lambda $ -invariant. By Lemma 2.20, there exists a definable $\Lambda $ -invariant $G_1\trianglelefteq G$ of finite index such that $G_1\subseteq C_G(H)$ . In particular, $G_1\cap H$ is central in $G_1$ . We fix such $G_1$ .

Assume that $G/H$ has an infinite $\Lambda $ -invariant definable abelian normal subgroup of the form $N/H$ for $N\trianglelefteq G$ . It follows that N is $\Lambda $ -invariant. Let $N_1:=N\cap G_1$ , an infinite normal subgroup of G of finite index in N and $H_1:=H\cap N_1$ , a central subgroup of $N_1$ . The quotient $N_1/H_1$ is isomorphic to $N_1H/H\subseteq N/H$ and so is abelian. Note that $N_1$ is also $\Lambda $ -invariant.

By Lemma 2.21 (2), $Z(N_1)$ has finite index in $N_1$ , and therefore, $\mathrm {dp\text {-}rk}(Z(N_1))=\mathrm {dp\text {-}rk}(N_1)=\mathrm {dp\text {-}rk}(N)=\mathrm {dp\text {-}rk}(N/H)$ . Because $N_1$ is $\Lambda $ -invariant and normal in G, so is $Z(N_1)$ . Hence, $Z(N_1)$ is a $\Lambda $ -invariant definable normal abelian subgroup of G of the same rank as $N_1/H$ . Clearly, if $N/H$ is B-definable for some $B\supseteq A$ , then so are $N_1$ and $Z(N_1)$ .

Proof. Since $\pi : K\to K/{\mathcal {O}}$ is a group homomorphism, and the image of a ball (centered at $0$ ) under $\pi $ is again a ball, it suffices to show that the claim is true for definable subgroups of $(K,+)$ . So let G be a subgroup of $(K,+)$ . Since $(K,+)$ is torsion-free, if G is finite, it is trivial. So we may assume G is infinite. Let B be the union of all sub-balls of G containing $0$ . If $B=K$ , then $G=K$ , and we are done, so assume $B\neq K$ . Because $\Gamma $ is definably complete, B is a ball itself, possibly $\{0\}$ . Since every infinite definable subset of K has an interior, and G is a group $B\neq \{0\}$ . We will show that $G=B$ .

Assume for contradiction that $G\neq B$ . In our settings, B is a divisible group (indeed, the maps $x\mapsto nx$ send B onto itself for all nonzero $n\in {\mathbb {N}}$ ), and since $(K,+)$ is torsion-free, it must be that $[G:B]=\infty $ . This means that G contains infinitely many disjoint maximal balls, cosets of B.

Assume that B is a closed ball. By the so-called (Cballs) property introduced in [Reference Halevi, Hasson and Peterzil12], which holds in our settings [Reference Halevi, Hasson and Peterzil12, Proposition 5.6, Lemma 5.10], only finitely many translates of B intersect G, so G contains only finitely many cosets of B, contradiction.

Assume then that B is open. After re-scaling G, we may assume that $B=\textbf {m}$ . Again, by (Cballs), G intersects only finitely many closed $0$ -balls. Consequently, ${\mathcal {O}}\cap G$ is an additive subgroup of K containing infinitely many cosets of $\textbf {m}$ . The image of ${\mathcal {O}}\cap G$ is, therefore, an infinite definable subgroup of $(\textbf {k},+)$ . However, under our assumptions, $\textbf {k}$ has no infinite definable proper subgroups, and thus, $G\cap {\mathcal {O}}={\mathcal {O}}$ contradicting the maximality of the ball $B=\textbf {m}$ . Thus, $G=B$ , with the desired conclusion.

Fact 3.3. Let $B_1,B_2\subseteq K$ be balls (possibly the whole of K).

1. (1) Every ball containing $1$ but not $0$ is a multiplicative subgroup of $K^\times $ .

2. (2) The point-set product $B_1\cdot B_2$ is also a ball.

3. (3) If $0\notin B_2$ , then their point-set quotient $B_1\cdot (B_2)^{-1}$ is also a ball.

Proof. We assume both $B_1$ and $B_2$ are not equal to K. The proof can be easily adapted to include this case as well.

(1) Well known.

(2) Let $B_1$ and $B_2$ be balls. It will suffice to show that $cB_1B_2$ is a ball for some $c\neq 0$ . So, as we proceed, we may freely replace $B_i$ with $cB_i$ for any such constant c.

Assume, first, that $0\in B_1$ but $0\notin B_2$ ; thus, $B_1B_2=\bigcup \{B_1b: b\in B_2\}$ is a chain of balls centered at $0$ . After multiplying by a suitable element, we may assume that $v(b)=0$ for all $b\in B_2$ and so $B_1b=B_1$ for all $b\in B_2$ , which gives $B_1B_2=B_1$ . If $0\in B_1\cap B_2$ , then after multiplying by suitable elements, we may assume that $B_1,B_2\in \{{\mathcal {O}},\textbf {m}\}$ ; in any of these cases, $B_1B_2$ is obviously a ball.

Assume, now, that $0\neq B_1\cup B_2$ . By multiplying by appropriate elements, we may assume that $1\in B_1\cap B_2$ , so both are multiplicative subgroups of $K^\times $ . Without loss of generality, $B_1\subseteq B_2$ . Then $B_2\subseteq B_1B_2\subseteq B_2B_2=B_2$ .

(3) If $0\notin B_2$ , then after possibly multiplying by an appropriate element, we get that $B_2$ is a multiplicative subgroup of $K^\times $ . Thus, $B_2^{-1}=B_2$ and (2) applies.

Proof. Since $I,J, H$ are definable subgroups of K, they are balls and so are their cosets, and because T is a group homomorphism, the image under T of a coset of I is also a coset of a subgroup, so viewed as a subset of K, it is a ball. Given $x\in H\setminus I$ , let

$$\begin{align*}S_x=\{w/z\in K:z\in x+I \wedge w\in T(x+I)\}.\end{align*}$$

As a quotient of two balls, $S_x$ is a ball, too (note that $0\notin x+I$ so Fact 3.3 applies). For $d\in K$ , let

$$\begin{align*}H_d=I\cup \{x\in H\setminus I :d\in S_x\}.\end{align*}$$

We claim that each $H_d$ is a subgroup of K (and when $I=0$ , possibly a singleton). To see this, let $H_d^{\prime }=\{x\in H\setminus I :d\in S_x\}$ ; by definition, $H_d^{\prime }\cap I=\emptyset $ . It follows directly from the definition of $H_d^{\prime }$ that if $x_1\in I$ and $x_2\in H_d^{\prime }$ , then $x_1\pm x_2\in H_d^{\prime }$ . So it remains to show that if $x_1,x_2\in H_d^{\prime }$ , then $x_1-x_2\in H_d$ . By assumption, $d\in S_{x_1}\cap S_{x_2}$ , so we can write, $d=w_1/z_1=w_2/z_2$ with $w_i\in T(x_i+I)$ and $z_i\in x_i+I$ . So $d(z_1-z_2)=w_1-w_2$ . If $z_1-z_2\in I$ , then $x_1+I=x_2+I$ , so obviously, $x_1-x_2\in H_d$ . Otherwise, $d=(w_1-w_2)/(z_1-z_2), \ z_1-z_2\in x_1-x_2+I$ and $w_1-w_2\in T(x_1+I)-T(x_2+I)=T(x_1-x_2+I)$ .

Hence, by Lemma 3.1, $H_d$ is a ball around $0$ . We use this fact now to show that the family $\{S_x:x\in K\}$ forms a chain of balls with respect to inclusion. Namely, we show that for $x_1,x_2\in H\setminus I$ , if $v(x_1)\leq v(x_2)$ , then $S_{x_1}\subseteq S_{x_2}$ . Let $d\in S_{x_1}$ . Since $H_d$ is a ball and $v(x_1)\leq v(x_2)$ , then $x_1\in H_d$ implies that $x_2\in H_d$ (i.e., $d\in S_{x_2}$ ).

Since V-minimal and power bounded T-convex valued fields are $1$ -h-minimal (see [Reference Cluckers, Halupczok and Rideau-Kikuchi6, Section 6]), they are definably spherically complete ([Reference Cluckers, Halupczok and Rideau-Kikuchi6, Lemma 2.7.1] – namely, the intersection of a definable chain of nonempty balls is nonempty. Thus, $\bigcap \limits _{x\in H\setminus I} S_x\neq \emptyset $ , and we let d be an element in the intersection.

Let $\hat H_d=\{ z\in H: d\cdot z\in T(z+I)\}$ . Since $T: H/I \to K/J$ is a homomorphism, $\hat H_d$ is a subgroup of $(K,+)$ . By definition, $H_d^{\prime }\subseteq \hat H_d$ and as both $\hat H_d$ and I are balls, either $I\subseteq \hat H_d$ or $\hat H_d\subseteq I$ . Since $H_d^{\prime }\cap I=\emptyset $ necessarily, $I\subseteq \hat H_d$ , and thus, $H_d\subseteq \hat H_d$ . However, by the choice of d, for all $x\in H\setminus I, \ d\in S_x$ , so $H=H_d=\hat H_d$ .

Finally, as $I\subseteq \hat H_d, \ d\cdot I\subseteq T(I)=J$ . Thus, $T(x+I)=d\cdot (x+I)+J$ for any $x\in H$ .

Proof. $(1)_1$ By Lemma 3.1, every definable subgroup of K is a ball, possibly trivial.

$(2)_1$ This is Lemma 3.4 for $I=\{0\}$ .

We now proceed with the induction step, assuming $(1)_{n-1}, (2)_{n-1}$ and prove $(1)_{n}$ :

Let $\pi :K^n\to K^{n-1}$ be the projection onto the first $n-1$ coordinates. By $(1)_{n-1}$ , we may assume that $\pi (H)=H_1\times \cdots \times H_{n-1}$ , for balls $H_i\subseteq K$ . Also, write $\ker (\pi )=H\cap (\{0\}^{n-1}\times K)$ as $\{0\}^{n-1}\times J$ , for a definable subgroup $J\subseteq K$ .

Notice that for every $(a,b), (a,c)\in H\subseteq K^{n-1}\times K$ , we have $b-c\in J$ , and hence, H can be viewed as the graph of a function $T:\pi (H) \to K/J$ , mapping a to $b+J$ ; that is,

$$\begin{align*}H=\{(a,b)\in K^n:a\in \pi(H)\, \wedge b\in T(a)\}. \end{align*}$$

By $(2)_{n-1}$ , there are $\alpha _1,\ldots , \alpha _{n-1}\in K$ , such that $T(x)=\sum _{i=1}^{n-1}\alpha _ix_i+J$ .

Hence,

$$\begin{align*}H=\left\{(x_1,\ldots, x_n)\in K^n: (x_1,\ldots, x_{n-1})\in \pi(H) \wedge x_n-\sum_{i=1}^{n-1} \alpha_i x_i\in J\right\}. \end{align*}$$

The groups J and $\alpha _iH_i$ , for $i=1,\ldots , n-1$ , are subgroups of $(K,+)$ , and hence, they are balls. Thus, for every $i=1,\ldots , n-1$ , either $J\subseteq \alpha _i H_i$ or $\alpha _i H_i\subseteq J$ . Note that if $\alpha _{i_0} H_{i_0}\subseteq J$ for some $i_0$ and $(x_1,\dots , x_{n-1})\in \pi (H)$ , then $x_n-\sum \limits _{i\neq i_0} \alpha _i x_i\in J$ iff $x_n-\sum \limits _i \alpha _ix_i\in J$ . So there is no harm assuming that $\alpha _i=0$ whenever $J\supseteq \alpha _i H_i$ and that $J\subseteq \alpha _i H_i$ whenever $\alpha _i\neq 0$ . Also, we may assume that for some $i, \ \alpha _i\neq 0$ , for otherwise, $H=\pi (H)\times J$ , and we are done.

Fix $\alpha _1, \dots , \alpha _{n-1}$ as above. Permuting the coordinates, if needed, we may assume that $v(\alpha _1)\leq v(\alpha _j)$ , for all $j=2,\ldots , n-1$ . Thus, we can write

$$\begin{align*}H=\left\{(x_1,\ldots, x_n): (x_1,\ldots, x_{n-1})\in \pi(H)\,\wedge \frac{1}{\alpha_1}x_n-(x_1+\sum_{i=2}^{n-1} \frac{\alpha_i}{\alpha_1} x_i)\in \frac{1}{\alpha_1}J\right\}.\end{align*}$$

Let $S(x_2,\ldots , ,x_n)=\frac {1}{\alpha _1}x_n-\sum _{i=2}^{n-1} \frac {\alpha _i}{\alpha _1}x_i.$ Then $S:K^{n-1}\to K$ is a linear map defined over ${\mathcal {O}}$ , and we have

(1) $$ \begin{align}H=\left\{(x_1,\ldots, x_n): (x_1,\ldots, x_{n-1})\in \pi(H)\,\wedge \, x_1 -S(x_2,\ldots, x_n) \in \frac{1}{\alpha_1}J\right\}.\end{align} $$

Let $\hat \pi (x_1,x_2,\ldots , x_n)=(x_2,\ldots , x_n)$ be the projection onto the last $n-1$ coordinates.

We get that

$$\begin{align*}H=\left\{(x_1,x_2,\ldots, x_n):(x_2,\ldots, x_n)\in \hat \pi(H) \,\wedge\, x_1-S(x_2,\ldots, x_n)\in \frac{1}{\alpha_1}J\right\}.\end{align*}$$

So $H\cap (K\times \{0\}^{n-1})=\frac {1}{\alpha _1}J\times \{0\}^{n-1}$ and, in particular, the map $(x_1,\dots , x_n)\mapsto x_1-S(x_2,\dots , x_n)$ from H to $\frac {1}{\alpha _1}J$ is surjective. We now define $F:K^n\to K^n$ by

$$\begin{align*}F(x_1,x_2,\ldots, x_n)=(x_1-S(x_2,\ldots, x_n),x_2, \ldots x_n).\end{align*}$$

Then F is over ${\mathcal {O}}$ , and by a direct computation, one sees that it has determinant $1$ ; hence, $F\in \operatorname {\mathrm {GL}}_n({\mathcal {O}})$ . It follows from the definition of F and the observation above that the restriction $F\restriction H$ is definable, injective and onto $\frac {1}{\alpha _1}J\times \hat \pi (H)$ .

By induction, there is $h\in \operatorname {\mathrm {GL}}_{n-1}({\mathcal {O}})$ such that $h(\hat \pi (H))$ is a product of balls. Hence, there is $g\in \operatorname {\mathrm {GL}}_n({\mathcal {O}})$ sending H to a product of balls. This ends the proof of $(1)_n$ .

For $(2)_n$ , we start with $T:H\to K/J$ . with $H\subseteq K^n$ , By $(1)_n$ , we may assume that $H=V_1\times \cdots \times V_n$ , for definable subgroups $V_i\subseteq K$ . Thus,

$$\begin{align*}T(x_1,\ldots, x_n)=T(x_1,0,\ldots, 0)+\cdots +T(0,\ldots, 0,x_n),\end{align*}$$

with all elements still in H. The result follows from the case $n=1$ .

Proof. (1) Consider $\hat H\subseteq K^n$ the preimage of H in $K^n.$ By Lemma 3.5, there is $g\in \operatorname {\mathrm {GL}}_n({\mathcal {O}})$ such that $g(\hat H) $ is a product of (possibly trivial) balls in K. Since $g\in \operatorname {\mathrm {GL}}_n({\mathcal {O}})$ , it descends to an automorphism of $(K/{\mathcal {O}})^n$ sending H to a product of balls in $(K/{\mathcal {O}})$ (possibly trivial ones).

For (2), we may assume that $H=V_1\times \cdots \times V_n$ for $V_i\subseteq K/{\mathcal {O}}$ and then

$$\begin{align*}T(x_1+{\mathcal{O}},\ldots, x_n+{\mathcal{O}})=T(x_1+{\mathcal{O}},0,\ldots,0)+\cdots+T(0,\ldots,0,x_n+{\mathcal{O}}),\end{align*}$$

with each element on the right inside H. We apply Lemma 3.4 with $I=J={\mathcal {O}}$ , so there are $d_1,\ldots , d_n\in {\mathcal {O}}$ (because $d_i{\mathcal {O}}\subseteq {\mathcal {O}}$ ), such that $T(x_1+{\mathcal {O}},\ldots , x_n+{\mathcal {O}})=d\cdot x_1+\cdots +d_n\cdot x_n+{\mathcal {O}}$ .

Proof. For the first part, we may think of T in coordinates and apply Lemma 3.5 $(2)_n$ to each coordinate map, obtaining $L\in \mathrm {End}((K/{\mathcal {O}})^n)$ extending T.

Assume now that T is injective, and we shall see that so is L. By Lemma 3.7 $(1)_n$ , after composing with a definable automorphism of $(K/{\mathcal {O}})^n$ , we may assume that $H=B_1\times \dots \times B_n$ , where each $B_i\subseteq K/{\mathcal {O}}$ is a ball around $0$ (possibly trivial).

Assume first that, for all $i, \ B_i$ is not the zero ball. If L, the extension of T provided above, were not injective, then, after permutation of the coordinates, we may assume the projection of $\ker (L)$ into $B_1$ is infinite. But then, $\ker (L)\cap B_1\times \{0_{n-1}\}$ is nontrivial, contradicting the injectivity of T.

So without loss of generality, we assume that $H=B_1\times \dots \times B_m\times \{0\}^{n-m}$ and that $B_i$ is nontrivial for $i\leq m$ . Since T is injective, $\mathrm {dp\text {-}rk}(T(H))=m=\mathrm {dp\text {-}rk}(H)$ , and hence, after a definable automorphism of $(K/{\mathcal {O}})^n$ (the range), we may assume that $T(H)=C_1\times \dots \times C_m\times \{0\}^{n-m}$ , where the $C_i\subseteq K/{\mathcal {O}}$ are balls with $r(C_i)<0$ (possibly $C_i=K/{\mathcal {O}}$ ). Setting $H_1=B_1\times \dots \times B_m$ and $H_2=C_1\times \dots \times C_m$ , the map T thus induces an injective isomorphism of $H_1$ and $H_2$ that, by what we have already noted, can be extended to a definable automorphism $L_1$ of $(K/{\mathcal {O}})^m$ .

Now, for $(x,y)\in (K/{\mathcal {O}})^m\times (K/{\mathcal {O}})^{n-m}$ , let $S(x,y)=(L_1(x),y)$ . This is an extension of T to an automorphism of $(K/{\mathcal {O}})^n$ .

Proof. By Lemma 3.7(2), there exist $L_1,L_2\in M_n({\mathcal {O}})$ such that for every $x\in K^n$ ,

$$ \begin{align*}f(x+{\mathcal{O}}^n)=L_1(x)+{\mathcal{O}}^n,\,\,\,\, f^{-1}(x+{\mathcal{O}}^n)=L_2(x)+{\mathcal{O}}^n.\end{align*} $$

It follows that for all $x\in K^n$ , we have $L_1\circ L_2(x)-x\in {\mathcal {O}}^n$ . It is easy to see that this forces the K-linear map $L_1\circ L_2(x)-x$ to be $0$ . Thus, $L_2=L_1^{-1}$ and both belong to $\operatorname {\mathrm {GL}}_n({\mathcal {O}})$ .

Proof. Assume everything is defined over some parameter set A and let p be a complete type over A which is concentrated on $H/{\mathcal {O}}$ with $\mathrm {dp\text {-}rk}(p)=1$ . As in [Reference Halevi, Hasson and Peterzil13, Section 3.2], there exists a unique complete type $\widehat p$ over A concentrated on H such that $\rho _*\widehat p=p$ . In particular, for any $a\models \widehat p$ , also $a+{\mathcal {O}}\models \widehat p$ .

By generic differentiability, $\widehat T$ and $\widehat T^{\prime }$ are both differentiable on $\widehat p$ (see [Reference Halevi, Hasson and Peterzil13, Lemma 3.17(1)]). A similar proof to that of [Reference Halevi, Hasson and Peterzil13, Lemma 3.17(2)] gives, for any $b\models \widehat p$ , that $\widehat T^{\prime }(b)\in B_J$ .

Proof.

The proof mimics [Reference Halevi, Hasson and Peterzil13, Lemma 3.17(3)]. Since there is one delicate adjustment toward the end, we give the whole argument. The reader may refer to [Reference Halevi, Hasson and Peterzil13, Section 3.2] for the relevant definitions and notions.

By saturation of ${\mathcal {K}}$ and the definition of $\widehat p$ , there exists a ball $B_0\subseteq \widehat p({\mathcal {K}})$ around a with $r(B_0)< \mathbb {Z}$ (see [Reference Halevi, Hasson and Peterzil13, Section 3.2]) and let $r_0:=r(B_0)$ . Note that $B_{>r_0+m}(a)\subseteq \widehat p({\mathcal {K}})$ for any natural number m.

By [Reference Halevi, Hasson and Peterzil13, Fact 3.13] applied to the function $\widehat T^{\prime }$ , there are an A-definable finite set C and $m\in {\mathbb {N}}$ such that ??†

(†) $$ \begin{align*} v(\widehat T^{\prime}(a)-\widehat T^{\prime}(x))=v((\widehat T^{\prime\prime}(a))+v(a-x) \end{align*} $$

for all x in any ball m-next to C around a, and $v(\widehat T^{\prime \prime }(x))$ is constant on that ball. By definition, the ball $B_{> r_0+m}(a)$ is contained in a ball m-next to C, so after possibly shrinking $B_0$ , we may assume that $v(\widehat T^{\prime \prime }(x))$ is constant on $B_0$ and that (†) holds on $B_0$ (see also [Reference Halevi, Hasson and Peterzil13, Lemma 3.14]).

If $\widehat T^{\prime \prime }(t)\equiv 0$ , the claim holds trivially. Otherwise, by [Reference Halevi, Hasson and Peterzil13, Fact 3.13], $\widehat T^{\prime }(B_{>r^{\prime }}(a))$ is an open ball of radius $v(\widehat T^{\prime \prime }(a))+r^{\prime }$ for any $r^{\prime }\geq r_0$ .

As $B_{>r_0}(a) \subseteq \widehat p(\widehat {\mathcal {K}})$ , we have $\widehat T^{\prime }(B_{>r_0}(a))\subseteq B_J$ . Since $J/{\mathcal {O}}$ is finite, we deduce that $v(\widehat T^{\prime \prime }(a))+r_0$ is either positive or a finite negative integer. Either way, for any $r^{\prime }>\mathbb {Z}$ satisfying that for any $n\in \mathbb {Z}, \ r^{\prime }-n>r_0$ , we get that $\widehat T^{\prime }(B_{>r^{\prime }}(a))$ is an open ball of radius $v(\widehat T^{\prime \prime }(a))+r^{\prime }>0$ .

So let r be such an element. Since $r/2$ also satisfies the same requirements, we deduce that $\widehat T^{\prime }(B_{>r/2}(a))$ is an open ball of radius $v(\widehat T^{\prime \prime }(a))+r/2>v(\widehat T^{\prime \prime }(a))+r>0$ .

We conclude that for any $b\in B:= B_{>r/2}(a), \ v(\widehat T^{\prime \prime }(b))+r>0$ .

Now, the proof of [Reference Halevi, Hasson and Peterzil13, Lemma 3.18] is applicable word-for-word, and we get that for every $a\models \widehat p$ , there is a ball $B, \ a\in B\subseteq \widehat p({\mathcal {K}})$ , such that for all $y\in B$ ,

$$\begin{align*}v(\widehat T(y)-\widehat T(a)-\widehat T^{\prime}(a)(y-a))>0. \end{align*}$$

Setting $c:=\widehat T^{\prime }(a)\in B_J$ , we get that for all $y\in B, \ \widehat T(y)-\widehat T(a)-c(y-a)\in \textbf {m} \subseteq J$ .

Let $U=B-a$ ; it is a subgroup of H. Let $x=y-a$ be an element of U (so $y\in B$ ). Since $\widehat T $ is a lift of a homomorphism, $\widehat T(x)+J=\widehat T(y)-\widehat T(a)+J=c(y-a)+J=cx+J$ .

Proof. (1) By [Reference Halevi, Hasson and Peterzil13, Lemmas 3.1(3), 3.10 (1)], the ball B contains all torsions points in $(K/{\mathcal {O}})^n$ . By [Reference Halevi, Hasson and Peterzil13, Lemma 3.10 (2)], N has nontrivial torsion. Thus, $N\cap B$ contains a nontrivial torsion point.

(2) Apply (1) to $N=\ker (f)$ .

Proof. Since $\mathrm {dp\text {-}rk}(N)=n-1$ , there exists a ball $B\subseteq N$ around $0$ . If there exists a coordinate projection $\pi $ and a natural number k for which $\pi \upharpoonright p^k (H\cap (B\times K/{\mathcal {O}}))$ is injective, then as $p^k(H\cap (B\times K/{\mathcal {O}}))=p^kH\cap (p^kB\times K/{\mathcal {O}})$ , we may apply Lemma 3.13 (2) and deduce that it is injective on $p^kH$ as well. Consequently, we may assume that $N=B=H_1\times \dots \times H_{n-1}$ is a product of balls.

Recall that $\rho :K\to K/{\mathcal {O}}$ is the quotient map. Since

$$\begin{align*}T(x_1,\ldots, x_{n-1})=T(x_1,0,\ldots, 0)+\cdots +T(0,\ldots, 0,x_{n-1}),\end{align*}$$

and denoting $\widehat T_i$ for a lift of $T(0,\dots ,x_i,\dots 0)$ to a partial map from K to K, we obtain

$$\begin{align*}\rho^{-1}(H)=\left\{(a_1,\dots,a_{n-1},a_n)\in K^n:(a_1,\dots,a_{n-1})\in \rho^{-1}(B)\, \wedge a_n+J= \sum_{i=1}^{n-1} \widehat T_i(a_i)+J\right\}.\end{align*}$$

Applying Lemma 3.12 to the $\widehat T_i$ , for each i, we find $c_i\in K$ and sub-balls $H_i^{\prime }\leq H_i$ such that $\widehat T_i(x)-c_ix\in J$ for elements of $H_i^{\prime }$ . Letting $B^{\prime }=H_1^{\prime }\times \dots \times H_{n-1}^{\prime }$ , we may, as above, replace B by $B^{\prime }$ and H by $H\cap (B^{\prime }\times K/{\mathcal {O}})$ . So we may assume that

$$\begin{align*}\rho^{-1}(H)=\left\{(a_1,\dots,a_{n-1},a_n)\in K^n:(a_1,\dots,a_{n-1})\in \rho^{-1}(B)\, \wedge a_n+J= \sum_{i=1}^{n-1} c_i\cdot a_i+J\right\}.\end{align*}$$

If $c_i=0$ for all $1\leq i\leq n-1$ , then H is equal to a product of $n-1$ balls together with $J/{\mathcal {O}}$ . If we choose $p^k$ large enough so that $p^kJ \subseteq {\mathcal {O}}$ , then $p^kH \subseteq (K/{\mathcal {O}})^{n-1}\times \{0\},$ and so projects injectively into the first $n-1$ coordinates.

We thus assume that $c_i\neq 0$ for some i. Setting $c_n=1$ , assume, without loss of generality, that $v(c_1)=\min _{1\leq i\leq n}\{v(c_i)\}$ .

Claim 3.14.1. $\rho ^{-1}(H)$ is equal to

$$\begin{align*}X:=\left\{(a_1,\dots,a_n)\in K^n:(a_2,\dots,a_n)\in P\, \wedge a_n+J= \sum_{i=1}^{n-1} c_i\cdot a_i+J\right\},\end{align*}$$

where P is the projection of $\widehat E$ on the last $n-1$ coordinates.

We get

$$\begin{align*}\rho^{-1}(H) =\left\{(a_1,\dots,a_n)\in K^n:(a_2,\dots,a_n)\in P\, \wedge a_1-\sum_{i=2}^n e_ia_i\in c_1^{-1}J\right\},\end{align*}$$

for some $e_i\in {\mathcal {O}}$ .

As $c_1^{-1}J/{\mathcal {O}}$ is finite as well, we can find some $k\in {\mathbb {N}}$ large enough so that $p^k(c_1^{-1}J)\subseteq {\mathcal {O}}$ . We claim that for any $(h_1,\dots ,h_n)\in p^kH, \ h_1$ is uniquely determined by $(h_2,\dots ,h_{n})$ . We will show that for a tuple in $\rho ^{-1}(p^kH)$ , the first coordinate is determined, up to ${\mathcal {O}}$ -equivalence, by the last $n-1$ coordinates.

To simplify the notation, we give the argument for $n=2$ , the general case is similar. Let $(a,b),(c,d)\in \rho ^{-1}(p^kH)$ , with $b-d\in {\mathcal {O}}$ . We want to prove that $a-c\in {\mathcal {O}}$ . As $\rho ^{-1}(p^kH)=p^k\widehat H+{\mathcal {O}}$ , we can write $(a,b)=(p^ka^{\prime }+o_1,p^kb^{\prime }+o_2)$ and $(c,d)=(p^kc^{\prime }+o_3,p^kd^{\prime }+o_4)$ , with $(a^{\prime },b^{\prime }),(c^{\prime },d^{\prime })\in \rho ^{-1}(H)$ .

We thus have $a^{\prime }-e_2(b^{\prime }+o_2), c^{\prime }-e_2(d^{\prime }+o_4) \in c_1^{-1}J$ . Since $p^k(c_1^{-1}J)\subseteq {\mathcal {O}}$ , we get that

$$\begin{align*}p^k(a^{\prime}-c^{\prime})-p^k(e_2(b^{\prime}-d^{\prime}))=p^k(a^{\prime}-c^{\prime})-e_2p^k(b^{\prime}-d^{\prime})\in {\mathcal{O}}.\end{align*}$$

By our assumption that $b-d\in {\mathcal {O}}$ (and since $p^k(b^{\prime }-d^{\prime })+{\mathcal {O}}=(b-d)+{\mathcal {O}}$ ), it follows that $p^k(b^{\prime }-d^{\prime })\in {\mathcal {O}}$ , and since $e_2 $ is assumed to be in ${\mathcal {O}}$ , it follows from the above that $p^k(a^{\prime }-c^{\prime })\in {\mathcal {O}}$ . By our assumptions, $p^k(a^{\prime }-c^{\prime })+{\mathcal {O}}=(a-c)+{\mathcal {O}}$ , and therefore, $a-c\in {\mathcal {O}}$ , as claimed.

Proof of Proposition 3.10.

We proceed by induction. The case $n=1$ is trivially true (take $k=0$ and $\pi _0=\operatorname {\mathrm {Id}}$ ).

Let $\pi :(K/{\mathcal {O}})^n\to (K/{\mathcal {O}})^{n-1}$ be the projection onto the first $n-1$ coordinates. We may assume that the kernel of this projection is finite: Indeed, let $H^i:=\ker (\pi ^i\upharpoonright H)$ for $\pi ^i$ the projection dropping the i-the coordinate. If all $H^i$ were infinite, then since $H\supseteq H^1\times \dots \times H^n$ , we would conclude that $\mathrm {dp\text {-}rk}(H)=n$ , and there is nothing to prove. Thus, we may assume that one of the $H^i$ is finite, and after permuting coordinates, assume that $i=n$ .

Write $\ker (\pi \upharpoonright H)$ as $\{0\}^{n-1}\times J$ , for a finite subgroup $J\subseteq K/{\mathcal {O}}$ . Since $\pi \upharpoonright H$ is finite-to-one, $\mathrm {dp\text {-}rk}(H)=\mathrm {dp\text {-}rk}(\pi (H))$ . Notice that for every $(a,b), (a,c)\in H\subseteq (K/{\mathcal {O}})^{n-1}\times K/{\mathcal {O}}$ , we have $b-c\in J$ , and hence, H can be viewed as the graph of a function $T:\pi (H) \to (K/{\mathcal {O}})/J$ , mapping a to $b+J$ ; that is,

$$\begin{align*}H=\{(a,b)\in (K/{\mathcal{O}})^n:a\in \pi(H)\, \wedge b+J= T(a)\}. \end{align*}$$

By the induction hypothesis applied to $\pi (H)\subseteq (K/{\mathcal {O}})^{n-1}$ , there exists $\ell \in \mathbb N$ , and a coordinate projection $\pi _1:(K/{\mathcal {O}})^{n-1}\to (K/{\mathcal {O}})^m$ such that $\pi _1\upharpoonright p^\ell \pi (H)$ is injective and $m=\mathrm {dp\text {-}rk}(\pi (H))$ . Without loss of generality, assume that $\pi _1$ is the projection onto the last m-coordinates $n-m,\dots , n-1$ . Let

$$\begin{align*}H_2= \{(a_1,\dots,a_{n-1},a_n)\in (K/{\mathcal{O}})^n:(a_1,\dots,a_{n-1})\in p^\ell \pi(H)\, \wedge a_n+J= T(a_1,\dots,a_{n-1})\},\end{align*}$$

and note that $p^\ell H\subseteq H_2$ .

By assumption, $H_2$ is definably isomorphic via $(\pi _1,\mathrm {id})$ to

$$\begin{align*}H_3= \{(a_{n-m},\dots,a_{n-1},a_n)\in (K/{\mathcal{O}})^{m+1}:(a_{n-m},\dots,a_{n-1})\in \pi_1(p^\ell \pi(H))\,\end{align*}$$

$$\begin{align*}\wedge a_n+J= S(a_{n-m},\dots,a_{n-1})\},\end{align*}$$

for $S=T\circ (\pi _1\upharpoonright p^\ell \pi (H))^{-1}$ .

Since $\mathrm {dp\text {-}rk}(\pi _1(p^\ell \pi (H)))=m$ , we may apply Lemma 3.14 to $H_3$ and find $r\in \mathbb {N}$ and a coordinate projection $\pi _2:(K/{\mathcal {O}})^{m+1}\to (K/{\mathcal {O}})^m$ (on some m coordinates) such that $\pi _2\upharpoonright p^rH_3$ is injective. As $H_2$ is isomorphic to $H_3$ via $(\pi _1,\mathrm {id})$ , by composing the coordinate projections, we get that $\pi _0=\pi _2\circ (\pi _1,id)$ is injective on $p^rH_2$ . Hence, it is also injective on $p^{r\ell }H\subseteq p^rH_2 $ .

Fact 4.1. Let ${\mathcal {M}}$ be a geometric structure and let E be a definable equivalence relation on $M^n$ . Then there exists a definable $S\subseteq M^n$ such that for every $x\in S, \ [x]\cap S$ is finite and $\dim (S)=dim(S/E)=\dim (M^n/E)$ .

Proof. We use the fact that, in our setting, the sort K is a geometric SW-uniformity. The proof relies on the following claim.

We now apply this claim to prove the statements of the corollary. First, let $Y_0$ be as in the statement; applying the claim for $Z=Y_0$ , we get $Y^{\prime }\subseteq Y_0$ strongly internal to K with $\mathrm {dp\text {-}rk}(Y^{\prime })=\dim (Y_0)$ . But since $Y_0$ is almost strongly internal to $K, \ \mathrm {dp\text {-}rk}$ and $\dim $ also coincide on $Y_0$ so $\mathrm {dp\text {-}rk}(Y^{\prime })=\dim (Y_0)=\mathrm {dp\text {-}rk}(Y_0)$ .

This result assures that the K-rank and the almost K-rank of Y are equal. To conclude, note that, since $\dim (Y)$ is obviously bounded below by the K-rank of Y, we only need to verify the other inequality. This is immediate by applying the claim to $Z=Y$ .

Proof. If D is an SW-uniformity, this follows from [Reference Simon and Walsberg31, Proposition 4.6], so we may assume that ${\mathcal {K}}$ is p-adically closed and D is either $\Gamma $ or $K/{\mathcal {O}}$ . If $D=\Gamma $ , this follows from cell-decomposition in Presburger arithmetic (as referred to in the proof of Fact 5.4). If $D=K/{\mathcal {O}}$ then by [Reference Halevi, Hasson and Peterzil13, Theorem 7.11(3)], there exists a definable subgroup $H\subseteq G$ with $\mathrm {dp\text {-}rk}(H)=n$ , the $K/{\mathcal {O}}$ -rank of G, definably isomorphic to a subgroup of $((K/{\mathcal {O}})^r,+)$ for some $r>0$ . By Proposition 3.10, we may assume, replacing H with a subgroup of the same dp-rank, that $r=n$ .

Proof. Let d denote the D-rank of $\operatorname {\mathrm {Fr}}(X)$ and let $X_1\subseteq \operatorname {\mathrm {Fr}}(X)$ be D-critical over A. Fix an A-generic $g\in X_1$ and $Y\ni g$ a definable basic open set. In particular, we can choose Y to be strongly internal to D.

By definition of $\operatorname {\mathrm {Fr}}(X)$ , it follows that $\operatorname {\mathrm {Fr}}(X)\cap Y= \operatorname {\mathrm {Fr}}(X\cap Y)$ . By Lemma 2.17, $\mathrm {dp\text {-}rk}(X_1\cap Y)=\mathrm {dp\text {-}rk}(X_1)$ . By the properties of SW-uniformities, ([Reference Simon and Walsberg31, Proposition 4.3, Lemma 2.3]), and since $X\cap Y$ can be identified with a subset of some $D^n, \ \mathrm {dp\text {-}rk}(\operatorname {\mathrm {Fr}}(X\cap Y))<\mathrm {dp\text {-}rk}(X\cap Y)$ . Thus, as $X_1\cap Y\subseteq \operatorname {\mathrm {Fr}}(X\cap Y)$ ,

$$\begin{align*}d=\mathrm{dp\text{-}rk}(X_1)\leq \mathrm{dp\text{-}rk}(\operatorname{\mathrm{Fr}}(X\cap Y))<\mathrm{dp\text{-}rk}(X\cap Y).\end{align*}$$

Since $X\cap Y$ is strongly internal to D (as Y was), its dp-rank is at most the D-rank of X, as needed.

Proof. Because G is a topological group, and a basis for the topology is definable, the closure of H, call it $H_1$ , is also a definable subgroup. Therefore, if $H_1\setminus H\neq \emptyset $ , then $H_1$ must contain a coset of H; thus, the D-rank of $H_1\setminus H$ is at least that of H contradicting Lemma 4.4. So H is closed in G.

Now, assume that the D-ranks of H and G are equal. This implies (by definition of $\nu _D$ ) that $\nu _D\vdash H$ . Since $\nu _D$ is open, and H is a group, this implies that H is open. Finally, as we have seen, if H is open, then it contains $\nu _D$ as a subgroup, and therefore, they have the same D-rank (since the D-rank of $\nu _D$ is maximal in G).

Proof. Let $g\in C_G(\nu _D)$ . By definition, $\nu _D\vdash C_G(g)$ , so by Lemma 4.5, $C_G(g)$ is a clopen subgroup of G. Now, $C_G(g)\cap X\neq \emptyset $ (as e is in the intersection), so definable connectedness of X implies $X\subseteq C_G(g)$ .

Proof. Assume that $X\subseteq S$ is infinite and $0$ -dimensional. By Fact 2.4, X (and hence also S) is locally almost strongly internal to some distinguished sort D. Namely, there is a definable infinite $X_1\subseteq X$ and a definable finite to one function $f:X_1\to f(X_1)\subseteq D^n$ . Since $\dim (X_1)\ge \dim (f(X_1))$ , necessarily $\dim (f(X_1))=0$ with $f(X_1)$ infinite. Hence, $D\neq K$ , so S is not K-pure.

For the converse, assume that S is not K-pure, witnessed by a definable infinite $X\subseteq S$ and a definable finite to one function $f:X\to D^n$ for some $D\neq K$ . Since $\dim (D)=0$ for $D\neq K$ , it follows that $\dim (f(X))=0$ , and hence, $\dim (X)=0$ . So, X is infinite and $0$ -dimensional.

Proof. A non-discrete locally Euclidean topological group is, by definition, a K-group, so (by Corollary 4.2) $\dim (G)>0$ , and since discrete groups are trivially locally Euclidean, we assume $\dim (G)>0$ . Since the topology is invariant under translations, it is sufficient to find a single $g\in G$ at which the topology is locally Euclidean. If $ n=\dim (G)>0$ , then, by Lemma 4.2, there exists a definable $X\subseteq G, \ \dim (X)=\dim (G)$ , such that X is strongly internal to K, over some A, and $\dim (G)$ is the K-rank of G. Given $g_1$ generic in X over A, it follows from Equation (2) at the beginning of Section 4.2 that there exists a definable $\tau _K$ -open set $U, \ g_1\in U\subseteq X$ , which is definably homeomorphic to an open set in $K^n$ .

Now, assume that $\tau _1,\tau _2$ are two locally Euclidean group topologies on G. Then for $g\in G$ , there are definable $U_1,U_2\ni g, \ U_i$ a $\tau _i$ -open set, and definable $f_i:U_i\to V_i\subseteq K^n$ , such that each $f_i$ is a homeomorphism between $U_i$ with the $\tau _i$ -topology and open $V_i$ with the $K^n$ -topology.

The map $f_2f_1^{-1}:f_1(U_1\cap U_2)\to V_2$ , is a definable injection. However, in SW-uniformities, definable bijections are homeomorphisms at generic points, [Reference Simon and Walsberg31, Corollary 3.8]. Thus, there is some $g_1\in U_1\cap U_2$ such that on $\tau _1, \tau _2$ open neighborhood of $g_1$ , the two topologies agree. Thus, $\tau _1=\tau _2$ .

Since Johnson’s admissible topology is locally Euclidean, the two topologies are the same.

Fact 5.1. For any A-definable set $X\subseteq (K/{\mathcal {O}})^n$ with $\mathrm {dp\text {-}rk}(X)=n$ and any A-generic $a\in X$ , there exists a ball $B\subseteq X$ with $a\in B$ .

Proof. If ${\mathcal {K}}$ is power-bounded T-convex or V-minimal, then this is [Reference Simon and Walsberg31, Corollary 2.7], and if ${\mathcal {K}}$ is p-adically closed, this is [Reference Halevi, Hasson and Peterzil13, Lemma 3.6].