Uniform Tits Alternative I: Introduction to heights

It is a classical theorem of Tits that a finitely generated linear group, i.e. one which embeds in $\mathrm{GL}_n(K)$ for some field $K$, can only be 'large' if it contains a non-abelian free group. More precisely, such a group is either virtually soluble or contains a non-abelian free group. This shows that linear groups are well-behaved in some sense: all their amenable subgroups are in fact elementary amenable. This is true for other large, important classes of groups as well:

• For hyperbolic groups, this is true by the ping-pong lemma. This is usually how you find free subgroups. 
• For mapping class groups of finite type surfaces this is due independently to McCarthy and Ivanov. Note however that this fails in general for infinite type surfaces. 
• For $\mathrm{Out}(F_n)$ this is due to Bestvina-Feighn-Handel, who proved this in two long and important papers. 
• For CAT(0) cubulated groups this is due to Sageev and Wise and comes from putting a lot of powerful theorems together.

Note that this fits the heuristic that $\mathrm{Out}(F_n)$, mapping class groups, and finitely generated linear groups should behave in very similar ways.
Tits' original proof uses a lot of algebraic machinery to find a ping-pong pair. Specifically, he uses arithmetic to exhibit a semisimple element with absolute value $>1$ for some clever choice of absolute value over the field of definition $K$, then studies the action on projective space over some completion $k$ of the original field and plays ping-pong in this space. Consequently, one learns a lot of mathematics when digesting this, and ultimately the proof isn't too difficult once one internalises sufficiently many theorems. In more recent history, Breuillard proved a uniform Tits alternative, which we will state and attempt to discuss the proof of. A subset $F$ in a group $G$ is said to be symmetric if $f \in F$ if and only if $f^{-1} \in F$. For a subset $F$ in a group $G$ and $n \in \mathbb{N}$ we denote by $F^{n}$ the set of elements which can be written as a product of at most $n$ elements in $F$.

Uniform Tits alternative: For every $d \in \mathbb{N}$ there is $N(d) \in \mathbb{N}$ such that for any field $K$ and $F$ a finite symmetric subset of $\mathrm{GL}_d(K)$ containing 1, either $F^{N(d)}$ contains two elements which freely generate a non-abelian free gorup or the group generated by $F$ is virtually soluble.

Breuillard's proof of this spans two hefty papers and is a marvel of technical achievement. Due to the length and complexity of the argument, we will only discuss parts of it and not give full details. In any case, it is questionable whether I would be able to give the details in any way meaningfully different to the original paper. The paper also lists multiple corollaries which are of interest to researchers in various fields of maths, but I leave these to the reader since I think the Uniform Tits alternative itself is the most exciting result in the paper.


The proof of the Uniform Tits alternative consists in reproducing Tits' proof almost
word by word while making sure that each step can be done in a uniform way.
The arithmetic step is much harder to perform, as we need a uniform gap $|\lambda |>1+\varepsilon ,$ where $\varepsilon $ is allowed to depend on $d$only. This first arithmetic step is to show a height gap theorem for non amenable linear groups.

Uniform Height Gap Theorem: There is a constant $\varepsilon =\varepsilon (d)>0$ such that if $F$ is a finite subset of $GL_{d}(\overline{\mathbb{Q}})$ generating a non amenable subgroup that acts strongly irreducibly, then $\widehat{h} (F)>\varepsilon $.

The key idea in the proof of this and the Tits alternative is to
introduce arithmetic heights in order to treat all absolute values of $K$ on
an equal footing. This first arithmetic step is needed only in
characteristic zero. In a second arithmetic step, Breuillard finds an absolute value
for which the geometric conditions needed for the ping-pong to work are
fulfilled by estimating the Arakelov heights of the characteristic subspaces of the matrices in $F$ in terms of the normalized height $\widehat{h}(F)$ and
by making use of the result, to be proved, which says that $\widehat{h}(F)$ can be realized up to a multiplicative factor as the height of some
conjugate of $F$ inside $SL_{d}(\overline{\mathbb{Q}})$. Once the right absolute value has been found, the actual geometric construction of the ping-pong pair follows Tits' geometric step very closely, with care taken to ensure that the estimates are uniform over all fields.

It will take us multiple posts to discuss a fraction of this, and our aim will be to showcase the breathtaking variety of ideas rather than provide a full proof. In this post we will define the height functions and discuss their geometric meaning.

First let us recall the statement of the Ping Pong lemma:

Let \( G \) be a group acting on a set \( X \) and let \( H_1, H_2, \ldots, H_k \) be subgroups of \( G \) where \( k \geq 2 \), such that at least one of these subgroups has order greater than 2. Suppose there exist pairwise disjoint nonempty subsets \( X_1, X_2, \ldots, X_k \) of \( X \) such that the following holds:

For any \( i \neq s \) and for any \( h \in H_i, h \neq 1 \), we have \( h(X_s) \subseteq X_i \).

Then \[\langle H_1, \dots, H_k \rangle = H_1 \ast \cdots \ast H_k.\]

Proof: By the definition of free product, it suffices to check that a given (nonempty) reduced word represents a nontrivial element of \( G \). Let \( w \) be such a word of length \( m \geq 2 \), and let
\[
w = \prod_{i=1}^{m} w_i,
\]
where \( w_i \in H_{\alpha_i} \) for some \( \alpha_i \in \{1, \dots, k\} \). Since \( w \) is reduced, we have \( \alpha_i \neq \alpha_{i+1} \) for any \( i = 1, \dots, m-1 \), and each \( w_i \) is distinct from the identity element of \( H_{\alpha_i} \).

We then let \( w \) act on an element of one of the sets \( X_i \). As we assume that at least one subgroup \( H_i \) has order at least 3, without loss of generality we may assume that \( H_1 \) has order at least 3.

We first make the assumption that \( \alpha_1 \) and \( \alpha_m \) are both 1 (which implies \( m \geq 3 \)). From here, we consider \( w \) acting on \( X_2 \). We get the following chain of containments:
\[
w(X_2) \subseteq \prod_{i=1}^{m-1} w_i(X_1) \subseteq \prod_{i=1}^{m-2} w_i(X_{\alpha_{m-1}}) \subseteq \cdots \subseteq w_1(X_{\alpha_2}) \subseteq X_1.
\]
By the assumption that different \( X_i \)'s are disjoint, we conclude that \( w \) acts nontrivially on some element of \( X_2 \), thus \( w \) represents a nontrivial element of \( G \).

To finish the proof we must consider the three cases:

• If \( \alpha_1 = 1, \alpha_m \neq 1 \), then let \( h \in H_1 \setminus \{ w_1^{-1}, 1 \} \) (such an \( h \) exists since by assumption \( H_1 \) has order at least 3);
•  If \( \alpha_1 \neq 1, \alpha_m = 1 \), then let \( h \in H_1 \setminus \{ w_m, 1 \} \);
• If \( \alpha_1 \neq 1, \alpha_m \neq 1 \), then let \( h \in H_1 \setminus \{1\} \).

In each case, \( hwh^{-1} \) after reduction becomes a reduced word with its first and last letter in \( H_1 \). Finally, \( hwh^{-1} \) represents a nontrivial element of \( G \), and so does \( w \). This proves the claim. $\blacksquare$

Let $G$ be a torsion-free hyperbolic group. It is well known that the centraliser of a cyclic subgroup is virtually cyclic, and a torsion-free virtually cyclic group is cyclic. If $G$ isn't cyclic, then there exists some $h \in G$ such that $gh \neq hg$. Then there exists some $M \geq 1$ such that for any $n, m \geq M, g^n$ and $h^m$ generate a non-abelian free group. To see this, consider the action of $g, h$ on the boundary of the group. If $g , h$ don't commute then the endpoints of their axes are all distinct and for sufficiently high powers of $g, h$ the action by these elements will send neighbourhoods of the endpoints in the boundary away from each other, so we can apply the ping pong lemma.

In our setting, the role of translation length will be a height function defined using the completions at all places of the field $\mathbb{\overline{Q}}$. In a later post we will discuss the geometric interpretation of this, but for today it will be mostly arithmetic.

Let $k$ be a local field of characteristic $0$. Let $\left\Vert \cdot \right\Vert _{k}$ be the standard norm on $k^{d},$ that is the canonical Euclidean (resp. Hermitian) norm if $k=\mathbb{R}$ (resp. $\mathbb{C}$) and the sup norm ($\left\Vert x\right\Vert _{k}=\max_{i}|x_{i}|_{k}$) if $k$ is non Archimedean. We will also denote by $\left\Vert \cdot \right\Vert _{k}$ the operator norm induced on the space of $d$ by $d$ matrices $M_{d}(k)$ by the standard norm $\left\Vert \cdot \right\Vert _{k}$ on $k^{d}.$ Let $Q$ be a bounded subset of matrices in $M_{d}(k)$. We set

\[\left\Vert Q\right\Vert _{k}=\sup_{g\in Q}\left\Vert g\right\Vert _{k} \]
and call it the norm of $Q$. Let $\overline{k}$ be an algebraic closure of $k.$ It is well known that the absolute value on $k$ extends to a unique absolute value on $\overline{k},$ hence the norm $\left\Vert \cdot \right\Vert _{k}$ also extends in a natural way to $\overline{k}^{d}$ and to $M_{d}(\overline{k}).$ This allows us to define the minimal norm of a bounded subset $Q$ of $M_{d}(k)$ as

\[E_{k}(Q)=\inf_{x\in GL_{d}(\overline{k})}\left\Vert xQx^{-1}\right\Vert _{k}\]
We will also need to consider the maximal eigenvalue of $Q,$ namely

\[\Lambda _{k}(Q)=\max \{|\lambda |_{k},\text{ }\lambda \in spec(q),q\in Q\} \]
where $spec(q)$ denotes the set of eigenvalues (the spectrum) of $q$ in $\overline{k}.$ We also set $Q^{n}=Q\cdot ...\cdot Q$ to be the set of all products of $n$ elements from $Q$. Finally, we introduce the spectral radius of $Q,$ that is

\[R_{k}(Q)=\lim_{n\rightarrow +\infty }\left\Vert Q^{n}\right\Vert _{k}^{\frac{1}{n}} \]
in which the limit exists (and coincides with $\inf_{n\in \mathbb{N}}\left\Vert Q^{n}\right\Vert _{k}^{\frac{1}{n}}$) because the sequence $\{\left\Vert Q^{n}\right\Vert _{k}\}_{n}$ is sub-multiplicative.

These quantities are related to one another. The key property concerning
them is given in the following result, which, together with its corollary
below (the list of properties), we call "spectral radius formula for several matrices" because of its parallel with the classical spectral radius formula relating the asymptotics of the powers of a matrix with its maximal eigenvalue:

Lemma 1: (Spectral Radius Formula for $Q$) Let $Q$ be a bounded subset of $M_{d}(k)$.

$(a)$ if $k$ is non Archimedean, there is an integer $q\in \lbrack 1,d^{2}]$
such that $\Lambda _{k}(Q^{q})=E_{k}(Q)^{q}.$

$(b)$ if $k$ is Archimedean, there is a constant $c=c(d)\in (0,1)$
independent of $Q$ and an integer $q\in \lbrack 1,d^{2}]$ such that $\Lambda
_{k}(Q^{q})\geq c^{q}\cdot E_{k}(Q)^{q}.$\newline


The proof, which we won't discuss in detail, makes use of two well-known theorems. Let $K$ be a field. The first is a theorem of Wedderburn that if an algebra $A$ over $K$ has a linear basis over $K$ consisting of nilpotent elements, then $A^{m}=0$ for some integer $m$. The second is a theorem of Engel that if $A$ is a subset of $M_{d}(K) $ such that $A^{m}=0$ for some integer $m,$ then $A$ can be simultaneously conjugated in $GL_{d}(K)$ inside $N_{d}(K),$ the subalgebra of upper triangular matrices with zeroes on and below the diagonal. Combined together, these facts yield:

Lemma 2 Let $K$ be a field. If $Q$ is any subset of $M_{d}(K)$ such
that $Q^{q}$ contains only nilpotent matrices for every $q,$ $1\leq q\leq
d^{2},$ then there is $g\in GL_{d}(K)$ such that $gQg^{-1}\subset N_{d}(K).$

Proof: Since $\dim _{K}M_{d}(K)\leq d^{2},$ the $K$-algebra generated by $Q$ has a
linear basis made of elements in $\cup _{1\leq q\leq d^{2}}Q^{q}$. By
Wedderburn and Engel, the result follows. $\blacksquare$


With the spectral radius formula at our disposal, we can now understand the relationships between the various quantities at hand, i.e. the minimal norm, spectral radius and maximal eigenvalue. 

These have the following properties:
Let $Q$ be a bounded subset of $M_{d}(k)$. We have

$(i)$ $\Lambda _{k}(Q)\leq R_{k}(Q)\leq E_{k}(Q)\leq \left\Vert Q\right\Vert
_{k},$ and $R_{k}(gQg^{-1})=R_{k}(Q)$ for any $g\in GL_{d}(\overline{k}),$

$(ii)$ $\Lambda _{k}(Q^{n})\geq \Lambda _{k}(Q)^{n}$, $E_{k}(Q^{n})\leq
E_{k}(Q)^{n}$ and $R_{k}(Q^{n})=R_{k}(Q)^{n}$ $\forall n\in \mathbb{N}$,

$(iii)$ $R_{k}(Q)=\lim_{n\rightarrow +\infty }E_{k}(Q^{n})^{\frac{1}{n}}=\inf_{n\in \mathbb{N}}E_{k}(Q^{n})^{\frac{1}{n}},$

$(iv)$ $R_{k}(Q)=\sup_{n\in \mathbb{N}}\Lambda _{k}(Q^{n})^{\frac{1}{n}},$

$(v)$ if $k$ is non Archimedean, $R_{k}(Q)=E_{k}(Q)$,

$(vi)$ if $k$ is Archimedean, $c\cdot E_{k}(Q)\leq R_{k}(Q)\leq E_{k}(Q)$,
where $c$ is the constant from Lemma 1 $(b)$.

Heights

Let $p$ be a prime number (abusing notation, we allow $p=\infty $). Fix an
algebraic closure $\overline{\mathbb{Q}_{p}}$ of the field of $p$-adic
numbers $\mathbb{Q}_{p}$ (if $p=\infty $, set $\mathbb{Q}_{p}=\mathbb{R}$).
We take the standard normalization of the absolute value on $\mathbb{Q}_{p}$
(i.e. $|p|_{p}=\frac{1}{p}$), while $|\cdot |_{\infty }$ is the standard absolute value on $\mathbb{R}$. It admits a unique extension to $\overline{\mathbb{Q}_{p}}$, which we again denote by $|\cdot |_{p}$. Let $\overline{\mathbb{Q}}$ be the field of all algebraic numbers over $\mathbb{Q}$ and $K$ a number field. Let $V_{K}$ be the set of equivalence classes of valuations on $K.$ For $v\in V_{K}$ let $K_{v}$ be the corresponding completion. 

For each $v\in V_{K},$ $K_{v}$ is a finite extension of $\mathbb{Q}_{p}$ for some prime $p$. We normalize the absolute value on $K_{v}$ to be the unique one which extends the standard abolute value on $\mathbb{Q}_{p}$. Equivalently $K_{v}$ has $n_{v}$ different embeddings in $\overline{\mathbb{Q}_{p}}$ and each of them gives rise to the same absolute value on $K_{v}.$ We identify $\overline{K_{v}}$, the algebraic closure of $K_{v}$ with  $\overline{\mathbb{Q}_{p}}$. Let $V_{f}$ be the set of finite places and $V_{\infty}$ the set of infinite places.

Let $d\in \mathbb{N}$ be an integer $d\geq 2$. For $v\in V_{K},$ in order
not to surcharge notation, we will use the subscript $v$ instead of $K_{v}$
in the quantities $E_{v}(F)=E_{K_{v}}(F)$, $\Lambda _{v}(F)=\Lambda
_{K_{v}}(F)$, etc.

Recall that if $x\in K$ then its height is by definition


\[h(x)=\frac{1}{[K:\mathbb{Q}]}\sum_{v\in V_{K}}n_{v}\log ^{+}|x|_{v}\]
It is well defined (i.e. independent of the choice of $K\ni x$). We will
make constant use of the following basic inequalities valid for every
algebraic numbers $x$ and $y$:\ $h(xy)\leq h(x)+h(y)$ and $h(x+y)\leq
h(x)+h(y)+\log 2.$

Let us similarly define the height of a matrix $f\in M_{d}(K)$ by

\[h(f)=\frac{1}{[K:\mathbb{Q}]}\sum_{v\in V_{K}}n_{v}\log ^{+}||f||_{v} \]

where $||f||_{v}$ is the operator norm of $f.$ We set the height of a finite
set $F$ of matrices in $M_{d}(K)$ to be

\[h(F)=\frac{1}{[K:\mathbb{Q}]}\sum_{v\in V_{K}}n_{v}\log ^{+}||F||_{v} \]
where $n_{v}=[K_{v}:\mathbb{Q}_{v}]$ and where $||F||_{v}=\max_{f\in
F}||f||_{v}.$ We also define the minimal height of $F$ as:


\[e(F)=\frac{1}{[K:\mathbb{Q}]}\sum_{v\in V_{K}}n_{v}\log ^{+}E_{v}(F)\]
and the normalized height of $F$ as:

\[\widehat{h}(F)=\frac{1}{[K:\mathbb{Q}]}\sum_{v\in V_{K}}n_{v}\log^{+}R_{v}(F)  \]
For any height $\mathbf{h}$ (i.e. $h,e$ or $\widehat{h}$), we also set $\mathbf{h}=\mathbf{h}_{\infty }+\mathbf{h}_{f}$, where $\mathbf{h}_{\infty }$
is the infinite part of $\mathbf{h}$ (i.e. the part of the sum over the infinite places of $K$) and $\mathbf{h}_{f}$ is the finite part of $\mathbf{h}$ (i.e. the part of the sum over the finite places of $K$). Note that these heights are well defined independently of the number field $K$ such that $F\subseteq M_{d}(K).$ We also set $h_{v}(F)=\log ^{+}||F||_{v}$ (resp. $e_{v}(F)=\log ^{+}E_{v}(F),$ etc) so that $h=\frac{1}{[K:\mathbb{Q}]}\sum_{v\in V_{K}}n_{v}h_{v},$ etc.


Remark 3 If we choose another basis of $\overline{\mathbb{Q}}^{d}$ the new height $h_{new}(F)$ differs only from the original height by a bounded additive error. Indeed there are only finitely many places where the new standard norm may differ from the original one. On the other hand $\widehat{h}(F)$ is independent of the choice of basis.


The above terminology is justified by the following facts:

Proposition 4: For any finite set $F$ in $M_{d}(\overline{\mathbb{Q}})$, we
have:

(a) $\widehat{h}(F)=\lim_{n\rightarrow +\infty }\frac{1}{n}h(F^{n})=\inf_{n\in \mathbb{N}}\frac{1}{n}h(F^{n}),$

(b) $e_{f}(F)=\widehat{h}_{f}(F)$ and $e(F)+\log c\leq \widehat{h}(F)\leq e(F)$ where $c$ is the constant in Lemma 1 $(b),$

(c) $\widehat{h}(F^{n})=n\cdot \widehat{h}(F)$ and $\widehat{h}(F\cup
\{Id\})=\widehat{h}(F),$

(d) $\widehat{h}(xFx^{-1})=\widehat{h}(F)$ if $x\in GL_{d}(\overline{\mathbb{Q}}).$


We also record the following:

Proposition 5:

1. $e(xFx^{-1})=e(F)$ for all $F$ finite in $M_{d}(\overline{\mathbb{Q}})$ and $x\in GL_{d}(\overline{\mathbb{Q}})$.
2. $e(F^{n})\leq n\cdot e(F),$
3. If $\lambda $ is an eigenvalue of an element of $F,$ then $h(\lambda)\leq \widehat{h}(F)\leq e(F),$
4.  If $F\subset GL_{d}(\overline{\mathbb{Q}})$ then $e(F\cup F^{-1})\leq (d|F|+d-1)\cdot e(F)$ and $e(F\cup \{1\})=e(F).$ If $F$ is a subset of $SL_{d}(\overline{\mathbb{Q}}),$ then $e(F\cup F^{-1})\leq (d-1)\cdot e(F)$.

Proof: The first three items are clear. For the last, observe that $||x^{-1}||_{v}=\frac{1}{|\det (x)|_{v}}||x||_{v}^{d-1}$ for any $x\in GL_{d}(K_{v})$ as can
be seen by expressing those norms in terms of the $KAK$ decomposition of $x$. Hence $||(F\cup F^{-1})||_{v}\leq ||F||_{v}^{d-1}\cdot \max \{\frac{1}{|\det (x)|_{v}},x\in F\cup \{1\}\}$ and $E_{v}(F\cup F^{-1})\leq
E_{v}(F)^{d-1}\cdot \max \{\frac{1}{|\det (x)|_{v}},x\in F\cup \{1\}\}.$ So $e(F\cup F^{-1})\leq (d-1)e(F)+\sum_{x\in F}h(\det (x)^{-1})$ i.e. $e(F\cup
F^{-1})\leq (d|F|+d-1)\cdot e(F).$ $\blacksquare$

We can also compare $e(F)$ and $\widehat{h}(F)$ when $\widehat{h}(F)$ is small:

Proposition 6: For every $\varepsilon >0$ there is $\delta =\delta
(d,\varepsilon )>0$ such that if $F$ is a finite subset of $SL_{d}(\overline{\mathbb{Q}})$ containing $1$ with $\widehat{h}(F)<\delta $, then $e(F)<\varepsilon $. Moreover $\widehat{h}(F)=0$ iff $e(F)=0.$

Corollary 7: Let $F$ be a finite subset of $GL_{d}(\overline{\mathbb{Q}}).$
Then $\widehat{h}(F)=0$ if and only if $e(F)=0.$


Proposition 8: (Height zero points) The height of a finite subset is 0 precisely when it generates a virtually unipotent subgroup, i.e. all eigenvalues of all elements are 1.


Proof: If $\widehat{h}(F)=0,$ then $e(F)=0$ by Corollary 7. Now by
Proposition 5, $e(F\cup F^{-1})=0,$ hence $\widehat{h}((F\cup F^{-1})^{n})=n\widehat{h}(F\cup F^{-1})=0$ for each $n\in \mathbb{N}$. Thus
every element from the group $\left\langle F\right\rangle $ generated by $F$ has only roots of unity as eigenvalues. However, it is known that $\left\langle F\right\rangle $ has a finite index subgroup $\Gamma _{0}$ for which no element has a non-trivial root of unity as eigenvalue. Therefore every element in $\Gamma _{0}$ must be unipotent, i.e. $\Gamma _{0}$ is unipotent. Conversely, if $\left\langle F\right\rangle $ is virtually unipotent, then every element in $\left\langle F\right\rangle $ has its eigenvalues among the roots of unity. In particular, as follows from the item 4 on the list of properties, $R_{v}(F)=1$ for every place $v.$ Hence $\widehat{h}(F)=0.$ $\blacksquare$

The above results dealt with small values of the normalized height. The next two results say that we have defined something which is 'coarsely' intrinsic to the group and isn't too sensitive to external things. The
following proposition says in substance that, provided $\left\langle
F\right\rangle $ has semisimple Zariski closure, the normalized height is
attained up to a constant by the height of some suitable conjugate of $F.$
We have

Proposition 9 (Comparison between $h$ and $\widehat{h}$) If $\mathbb{G}$ is a semisimple algebraic group over $\overline{\mathbb{Q}}$ and $(\rho ,V)$ a finite dimensional linear representation of $\mathbb{G}$, then there is $C\geq 1$ and there is a choice of a basis on $V$ with associated height function $h$ on $End(V),$ such that if $F$ is any finite subset generating a Zariski-dense subgroup, we have

\[\widehat{h}(\rho (F))\leq e(\rho (F))\leq h(\rho (gFg^{-1}))\leq C\cdot \widehat{h}(\rho (F)) \]
 

for some $g\in \mathbb{G}(\overline{\mathbb{Q}}).$

Recall from Remark 3 that if we change the basis of $V$ the associated height differs from the original one only by an additive constant. It is important for the applications as it allows us to conjugate $F$ back in the \textquotedblleft right position\textquotedblright. Observe that by definition $e(F)$ is equal to the infimum of $h(gFg^{-1})$ when $g=(g_{v})_{v\in V_{K}}$ is allowed to vary among the full group of ad \`{e}les $GL_{d}(\mathbb{A}).$ This proposition shows that this infimum is attained up to a multiplicative constant on principal ad\`{e}les, i.e. on $GL_{d}(\overline{\mathbb{Q}}).$ The condition that the Zariski closure of
the group generated by $F$ should be semisimple is important as easy examples show that the result of the proposition can fail if for instance $F$ normalizes a unipotent subgroup.

The normalized height $\widehat{h}$ was defined for an arbitrary finite
subset of $GL_{d}(\overline{\mathbb{Q}})$. If $\mathbb{G}$ is an arbitrary
semisimple group, one can define the normalized height for $\mathbb{G}$ as
the one you obtain after taking some absolutely irreducible representation
of $\mathbb{G}$ which is non trivial on each factor of $\mathbb{G}$. The
following proposition shows that up to constants, this height is independent
of the choice of the representation.

Proposition 10: Invariance under change of representation) Let $\mathbb{G}$ be a semisimple algebraic group over $\overline{\mathbb{Q}}$ and $(\rho _{i},V_{i})$ for $i=1,2$ be two finite dimensional linear representations of $\mathbb{G}$ which are non trivial on each simple factor of $\mathbb{G}$. Let $h_{i}$ be a height function on $End(V_{i})$ defined as above by the choice of a basis in each $V_{i}.$ Then there are constants $C_{12},C_{12}^{\prime }\geq 1$ such that for any finite subset $F$ of $\mathbb{G}(\overline{\mathbb{Q}}),$ we have

\[ \frac{1}{C_{12}}\cdot h_{2}(\rho _{2}(F))-C_{12}^{\prime }\leq h_{1}(\rho
_{1}(F))\leq C_{12}\cdot h_{2}(\rho _{2}(F))+C_{12}^{\prime } \]
In particular 

\[\frac{1}{C_{12}}\cdot \widehat{h_{2}}(\rho _{2}(F))\leq \widehat{h_{1}}(\rho
_{1}(F))\leq C_{12}\cdot \widehat{h_{2}}(\rho _{2}(F)) \]
Moreover the constant $C_{12}$ depends only on $\rho _{1}$ and $\rho _{2}$
and is independent of the choice of basis used to define $h_{1}$ and $h_{2}.$