Dense images of the power maps for a disconnected real algebraic group

Theorem 1.1

Let 𝐺 be a complex algebraic group, not necessarily Zariski-connected, defined over ℝ. Let 𝐴 be a subgroup of 𝐺 with G ⁢ ( R ) * ⊂ A ⊂ G ⁢ ( R ) and k ∈ N . Then the following are equivalent.

1. P k ⁢ ( A ) is dense in 𝐴.

2. ( k , ∘ ( A / G ( R ) * ) ) = 1 , and P k ⁢ ( G ⁢ ( R ) * ) is dense.

3. ( k , ∘ ( A / G ( R ) * ) ) = 1 , and P k : S ⁢ ( G ⁢ ( R ) * ) → S ⁢ ( G ⁢ ( R ) * ) is surjective.

4. P k : S ⁢ ( A ) → S ⁢ ( A ) is surjective.

In particular, if 𝐺 is a Zariski-connected algebraic group defined over ℝ, then for an odd k ∈ N , P k ⁢ ( G ⁢ ( R ) ) is dense in G ⁢ ( R ) if and only if P k ⁢ ( G ⁢ ( R ) * ) is dense in G ⁢ ( R ) * .

Corollary 1.2

Let 𝐺 be an algebraic group defined over ℝ, which is not necessarily Zariski-connected. Let k ∈ N . Then P k : G ⁢ ( R ) → G ⁢ ( R ) is surjective if and only if P k ⁢ ( Z G ⁢ ( R ) ⁢ ( u ) ) is dense in Z G ⁢ ( R ) ⁢ ( u ) for any unipotent element u ∈ G ⁢ ( R ) .

Theorem 2.1

Theorem 2.1 ([14, Theorem 8.9 (c)])

Let 𝐺 be a complex algebraic group, which is not necessarily connected. Suppose the connected component G ∘ is reductive. Then, for each a ∈ G , the set of semisimple elements S ⁢ ( G ) ∩ G ∘ ⁢ a is a non-empty Zariski-open subset of the coset G ∘ ⁢ a . In particular, S ⁢ ( G ) is Zariski-dense in 𝐺.

Proposition 2.2

Let 𝐺 be a complex algebraic group, not necessarily Zariski-connected, defined over ℝ such that G 0 is reductive. Then S ⁢ ( G ⁢ ( R ) ) contains an open dense subset of G ⁢ ( R ) in the Hausdorff topology of G ⁢ ( R ) .

Proof

Since G ⁢ ( R ) / G 0 ⁢ ( R ) embeds in G / G 0 , it follows that G ⁢ ( R ) / G 0 ⁢ ( R ) is a finite group. Let g 1 , … , g p ∈ G ⁢ ( R ) such that

It is enough to prove that, for all i = 1 , … , p , the set S ⁢ ( g i ⁢ G 0 ⁢ ( R ) ) contains an open dense subset of g i ⁢ G 0 ⁢ ( R ) in the Hausdorff topology of G ⁢ ( R ) . By Theorem 2.1, there is a Zariski-open set 𝑊 of g i ⁢ G 0 consisting of semisimple elements of 𝐺.

Let X sm be the set of smooth points of an irreducible affine variety 𝑋 defined over ℝ, and M := X sm ⁢ ( R ) ≠ ∅ . Let W ≠ ∅ be a Zariski-open set of 𝑋. Then W ∩ M is an open dense subset of 𝑀 in the Hausdorff topology of 𝑀.

Now observe that g i ⁢ G 0 sm ⁢ ( R ) = g i ⁢ G 0 ⁢ ( R ) . Hence, by the above fact, it follows that W ∩ g i ⁢ G 0 ⁢ ( R ) is open dense in g i ⁢ G 0 ⁢ ( R ) . ∎

Remark 2.3

Let 𝐺 be a complex algebraic group defined over ℝ. Then any element g ∈ G ⁢ ( R ) can be written as g = g s ⁢ g u , where g s , g u ∈ G ⁢ ( R ) and g s is semisimple, g u is unipotent and g , g s , g u commute with each other. Then we have Ad g = Ad g s ⁢ Ad g u . Since all the eigenvalues of Ad g u are 1, the generalized eigenspace for eigenvalue 1 of Ad g is the same as that of Ad g s . Note that, as Ad g s is semisimple, N ⁢ ( Ad g s - I ) = ker ⁡ ( Ad g s - 1 ) . The dimension of ker ⁡ ( Ad g s - 1 ) is equal to the dimension of the centralizer Z G ⁢ ( R ) ⁢ ( g s ) . So 𝑔 is regular in G ⁢ ( R ) if and only if Z G ⁢ ( R ) ⁢ ( g s ) has minimal dimension. In particular, 𝑔 is regular in G ⁢ ( R ) if and only if g s is regular in G ⁢ ( R ) .

Remark 2.4

The definition of regular element, mentioned above, coincides with the definition of regular element in [3, § 12.2] for the algebraic case. In the literature, there is another notion of regular element of an algebraic group (see [20]). Let 𝐺 be an algebraic group over an algebraically closed field. An element g ∈ G is said to be regular if dim ⁡ Z G ⁢ ( g ) ≤ dim ⁡ Z G ⁢ ( x ) for all x ∈ G . Note that, according to this definition, a unipotent element may be a regular element; in fact, if 𝐺 is connected and semisimple, then it admits a regular unipotent element (see [20]). However, our notion of regular element is different from that. For a semisimple group 𝐺 as above, it is easy to see that unipotent elements are not regular in our sense. Indeed, let 𝑢 be a unipotent element of a semisimple algebraic group 𝐺; then Ad u is unipotent, and hence the nilspace N ⁢ ( Ad u - I ) is the whole space, which is not of minimal dimension.

Lemma 2.5

Let 𝐺 be a complex algebraic group, not necessarily Zariski-connected, defined over ℝ. Let k ∈ N . Then the following are equivalent.

1. P k ⁢ ( S ⁢ ( Reg ⁢ ( G ⁢ ( R ) * ) ) ) ⊃ S ⁢ ( Reg ⁢ ( G ⁢ ( R ) * ) ) .

2. P k : S ⁢ ( G ⁢ ( R ) * ) → S ⁢ ( G ⁢ ( R ) * ) is surjective.

3. The image of the map P k : G ⁢ ( R ) * → G ⁢ ( R ) * is dense.

Proof

First, we will show that each of (1) and (2) implies ( 3 ) . In view of [1, Theorem 1.1], it is enough to show that

Let g ∈ Reg ⁢ ( G ⁢ ( R ) * ) . Let g = g s ⁢ g u , where g s and g u are respectively the semisimple part and unipotent part of the Jordan decomposition of 𝑔. By Remark 2.3, 𝑔 is regular if and only if g s is regular. Hence, either (1) or (2) implies that there exists h s ∈ S ⁢ ( Reg ⁢ ( G ⁢ ( R ) * ) ) ( h s ∈ S ⁢ ( G ⁢ ( R ) * ) ) such that h s k = g s . Since g u is unipotent, there exists a unique unipotent element h u such that h u k = g u . As the Zariski-closure of the cyclic subgroups generated by g u and h u are the same, g s commutes with h u .

Since h u is unipotent, there exists a unique nilpotent element X ∈ Lie ⁢ ( G ⁢ ( R ) * ) such that h u = exp ⁡ X . As g s commutes with h u , we have g s ⁢ exp ⁡ X ⁢ g s - 1 = exp ⁡ X , which gives

Now, by the uniqueness of the element X ∈ Lie ⁢ ( G ⁢ ( R ) * ) , we get Ad g s ⁢ ( X ) = X . Therefore, Ad h s k ⁢ ( X ) = X . By [1, Lemma 2.1], h s is regular and P k -regular. This implies Ad h s ⁢ ( X ) = X , which in turn shows that h s commutes with h u . Hence, we have g = g s ⁢ g u = h s k ⁢ h u k = ( h s ⁢ h u ) k .

Now suppose that ( 3 ) holds. Let g ∈ S ⁢ ( G ⁢ ( R ) * ) or S ⁢ ( Reg ⁢ ( G ⁢ ( R ) * ) ) . Then there exists a Cartan subgroup C ⊂ G ⁢ ( R ) * such that g ∈ C . By [1, Theorem 1.1], we have P k ⁢ ( C ) = C . Thus there exists h ∈ C such that h k = g . Since 𝑔 is semisimple (regular), ℎ is also semisimple (regular), which proves the lemma. ∎

Proposition 2.6

Let 𝐺 be a complex reductive algebraic group, not necessarily Zariski-connected, defined over ℝ. Let k ∈ N . Then the following are equivalent.

1. The image of P k : G ⁢ ( R ) → G ⁢ ( R ) is dense.

2. The map P k : G ⁢ ( R ) / G ⁢ ( R ) * → G ⁢ ( R ) / G ⁢ ( R ) * is surjective and the image of P k : G ⁢ ( R ) * → G ⁢ ( R ) * is dense.

3. The map P k : G ⁢ ( R ) / G 0 ⁢ ( R ) → G ⁢ ( R ) / G 0 ⁢ ( R ) is surjective and the image of P k : G 0 ⁢ ( R ) → G 0 ⁢ ( R ) is dense.

Proof

This follows by applying [7, Theorem 5.5] and Lemma 2.5, so we omit the details. ∎

Lemma 3.1

Let 𝐺 be a complex reductive algebraic group, which is not necessarily Zariski-connected, defined over ℝ. Let k ∈ N . Let G ⁢ ( R ) * ⊂ A ⊂ G ⁢ ( R ) . Suppose that P k ⁢ ( A ) is dense in 𝐴. Then, for any s ∈ S ⁢ ( A ) , there exists a closed abelian subgroup B s (containing 𝑠) of 𝐴 with finitely many connected components such that P k : B s → B s is surjective.

Proof

In view of Propositions 2.2 and 2.6, we note that P k ⁢ ( A ) is dense in 𝐴 if and only if P k : S ⁢ ( A ) → S ⁢ ( A ) is surjective. By [7, Theorem 5.5], P k : S ⁢ ( A ) → S ⁢ ( A ) is surjective if and only if 𝑘 is co-prime to the order of A / G ⁢ ( R ) * and

is surjective. We conclude our assertion by following the arguments of the proof of [7, Theorem 5.5 (1)].

Let s ∈ S ⁢ ( A ) . If the order of A / G ⁢ ( R ) * is 𝑚, then s m ∈ G ⁢ ( R ) * . In [7, Theorem 5.5, p. 230], it is shown that there exists a maximal ℝ-torus 𝑇 of G ′ = Z G ⁢ ( s m ) such that s ⁢ T ⁢ s - 1 = T and P k : Z T ⁢ ( R ) * ⁢ ( s ) → Z T ⁢ ( R ) * ⁢ ( s ) is surjective. Moreover, it is shown in [7, Theorem 5.5, Case 1 and Case 2] that s = ( t c ⁢ s d ) k for some c , d ∈ Z , where t ∈ Z T ⁢ ( R ) * ⁢ ( s ) . Also, s n ∈ Z T ⁢ ( R ) * ⁢ ( s ) for some positive integer 𝑛.

Now we consider the subgroup 𝐻 of 𝐴 generated by Z T ⁢ ( R ) * ⁢ ( s ) and the element 𝑠. Then 𝐻 is a closed abelian subgroup of 𝐴 and H / H * is finite. Further, P k : H → H is surjective as P k : Z T ⁢ ( R ) * ⁢ ( s ) → Z T ⁢ ( R ) * ⁢ ( s ) is surjective and s = ( t c ⁢ s d ) k ∈ H . Now set B s := H . ∎

Theorem 3.2

Theorem 3.2 ([11])

Let 𝐺, 𝑁, 𝐴 be as above. Let N = N 0 ⊃ N 1 ⊃ ⋯ ⊃ N r = { e } be a series of closed connected normal subgroups N j of 𝑁 as above such that the action of A = G / N on the vector space V j = N j / N j + 1 (induced by conjugation) is irreducible for all j ∈ { 0 , 1 , … , r - 1 } . Let x ∈ G and a = x ⁢ N ∈ A . Suppose there exists b ∈ A with b k = a such that, for any V j , if the action of 𝑎 on V j is trivial, then the action of 𝑏 on V j is trivial. Then, for n ∈ N , there exists y ∈ G such that y ⁢ N = b and y k = x ⁢ n .

Proof of Theorem 1.1

(1) ⇒ (2): This is obvious.

(2) ⇒ (1): We first give a brief outline of the proof and later explain the steps. In step 1, we assume G ⁢ ( R ) = L ⁢ ( R ) ⋉ R u ⁢ ( G ) ⁢ ( R ) for some Levi subgroup 𝐿 (of 𝐺) defined over ℝ and consider a dense set D = S ⁢ ( A ∩ L ⁢ ( R ) ) ⁢ R u ⁢ ( G ) ⁢ ( R ) of 𝐴. By using the hypothesis, we see that P k : S ⁢ ( A ∩ L ⁢ ( R ) ) → S ⁢ ( A ∩ L ⁢ ( R ) ) is surjective, i.e., P k ⁢ ( A ∩ L ⁢ ( R ) ) is dense. In step 2, to each element x 0 ∈ S ⁢ ( A ∩ L ⁢ ( R ) ) , we associate a subgroup B x 0 (containing x 0 ) of 𝐴 such that B x 0 is abelian with finitely many components and P k : B x 0 → B x 0 is surjective. Then we construct a solvable subgroup G x 0 of 𝐴 given by G x 0 = B x 0 ⋉ R u ⁢ ( G ) ⁢ ( R ) . We show there exists a dense open set W x 0 of G x 0 such that each element of W x 0 has a 𝑘-th root in G x 0 (and hence in 𝐴). In step 3, we show the existence of a dense set 𝑊 (of 𝐴) whose elements have 𝑘-th roots in 𝐴 as required.

Step 1. Since S ⁢ ( L ⁢ ( R ) ) is dense in L ⁢ ( R ) by Proposition 2.2, and A ∩ L ⁢ ( R ) is open in L ⁢ ( R ) , S ⁢ ( A ∩ L ⁢ ( R ) ) is dense in A ∩ L ⁢ ( R ) . Since G ⁢ ( R ) * ⊂ A ⊂ G ⁢ ( R ) , we have L ⁢ ( R ) * ⊂ A ∩ L ⁢ ( R ) ⊂ L ⁢ ( R ) . Note that if ( k , ∘ ( A / G ( R ) * ) ) = 1 , then ( k , ∘ ( ( A ∩ L ( R ) ) / L ( R ) * ) ) = 1 . Also, P k ⁢ ( G ⁢ ( R ) * ) is dense in G ⁢ ( R ) * , which implies that P k ⁢ ( L ⁢ ( R ) * ) is dense in L ⁢ ( R ) * . Now, by Lemma 2.5 applied to the reductive group L ⁢ ( R ) * , we have that P k : S ⁢ ( L ⁢ ( R ) * ) → S ⁢ ( L ⁢ ( R ) * ) is surjective. So P k : S ⁢ ( A ∩ L ⁢ ( R ) ) → S ⁢ ( A ∩ L ⁢ ( R ) ) is surjective by [7, Theorem 5.5]. Hence, P k ⁢ ( A ∩ L ⁢ ( R ) ) is dense in A ∩ L ⁢ ( R ) .

Step 2. For a given x 0 ∈ S ⁢ ( A ∩ L ⁢ ( R ) ) , by Lemma 3.1, there exists a closed abelian subgroup B x 0 of A ∩ L ⁢ ( R ) containing x 0 and P k ⁢ ( B x 0 ) = B x 0 . Also, B x 0 has finitely many connected components in the real topology. Consider the subgroup G x 0 = B x 0 ⋉ R u ⁢ ( G ) ⁢ ( R ) of 𝐴. Note that the unipotent radical N = R u ⁢ ( G ) ⁢ ( R ) is a simply connected nilpotent Lie group. Let

be the series as in Theorem 3.2. Let V j := N j / N j + 1 for j = 0 , 1 , … , r - 1 . As 𝑁 is simply connected, V j ’s are all finite-dimensional real vector spaces, and the B x 0 -actions on the V j ’s are irreducible.

Now, for j = 0 , 1 , … , r - 1 , let F j be the set of elements of B x 0 which act trivially on V j under the conjugation action. Note that F j is a closed subgroup of B x 0 for all 𝑗. We shall consider the following two cases separately:

1. dim ⁡ ( F j ) < dim ⁡ ( B x 0 ) for all 𝑗,

2. for some 𝑗, dim ⁡ ( F j ) = dim ⁡ ( B x 0 ) .

Suppose (i) holds. Then U ′ := B x 0 - ⋃ j F j is a dense open set in B x 0 . Therefore, all elements in U ′ act non-trivially on all the V j ’s, and hence, by Theorem 3.2, g ⁢ N ⊂ P k ⁢ ( G x 0 ) for all g ∈ U ′ . If we take W x 0 = U ′ × N , then W x 0 is a dense open subset of G x 0 such that W x 0 ⊂ P k ⁢ ( G x 0 ) .

Now suppose (ii) holds. Let I ⊂ { 0 , 1 , … , r - 1 } be the set of indices such that dim ⁡ ( F j ) = dim ⁡ ( B x 0 ) for all j ∈ I . Then ⋃ i ∉ I F i is a proper closed analytic subset of B x 0 of smaller dimension. Note that B x 0 ∖ ⋃ i ∉ I F i is a dense open set in B x 0 .

If dim ⁡ ( F j ) = dim ⁡ ( B x 0 ) , then F j is the union of some connected components of B x 0 (containing the identity component of B x 0 ). Fix j ∈ I . We will show that, for any x ∈ F j ∖ ⋃ i ∉ I F i , the coset x ⁢ N ⊂ P k ⁢ ( G x 0 ) .

It might happen that 𝑥 belongs to F s (for s ∈ I ) other than F j . Let

Clearly, j ∈ J and x ∈ ⋂ i ∈ J F i ⊂ F j . To prove the coset x ⁢ N has a 𝑘-th root, we will apply Theorem 3.2, and for that, we will show the existence of a 𝑘-th root of 𝑥 in ⋂ i ∈ J F i .

Since P k ⁢ ( B x 0 ) = B x 0 , we get that P k : B x 0 / B x 0 * → B x 0 / B x 0 * is surjective, and hence 𝑘 is co-prime to the order of the component group Γ = B x 0 / B x 0 * . Let the cardinality of Γ be 𝑛. Then there exist integers 𝑎 and 𝑏 such that a ⁢ k + b ⁢ n = 1 . Without loss of generality, we can choose 𝑎 to be a positive integer. Note that x = x a ⁢ k + b ⁢ n = ( x a ) k ⁢ ( x n ) b , where x n ∈ B x 0 * , and so does ( x n ) b . As B x 0 * is a connected abelian group, it is divisible. So there exists x ′ ∈ B x 0 * such that x ′ ⁣ k = ( x n ) b . This gives x = ( x a ) k ⁢ x ′ ⁣ k = ( x a ⁢ x ′ ) k as both x a and x ′ commute with each other. Clearly, we have x a ⁢ x ′ ∈ ⋂ i ∈ J F i ⊂ F j . This implies that, for l ∈ { 0 , 1 , … , r - 1 } , whenever the action of 𝑥 on V l is trivial, there exists a 𝑘-th root of 𝑥 (namely x a ⁢ x ′ ) which acts trivially on V l . Therefore, by Theorem 3.2, we conclude that x ⁢ n ∈ P k ⁢ ( G x 0 ) for all n ∈ N .

Finally, since j ∈ I is arbitrary, this proves the claim. Hence, there exists a dense open set W ′ in B x 0 such that W x 0 ′ ⊂ P k ⁢ ( G x 0 ) , where W x 0 ′ = W ′ × N . Thus, for both cases, we have a dense open set W x 0 in G x 0 such that each element of W x 0 has a 𝑘-th root in G x 0 and hence in 𝐴.

Step 3. We note that S ⁢ ( A ∩ L ⁢ ( R ) ) ⁢ N = ⋃ x 0 ∈ S ⁢ ( A ∩ L ⁢ ( R ) ) G x 0 . Now set

Then 𝑊 is a dense set in 𝐴 such that each element of 𝑊 has a 𝑘-th root in 𝐴, which proves (2) implies (1).

(2) ⇔ (3) follows from Lemma 2.5.

(3) ⇔ (4) follows from [7, Theorem 5.5].

Borel and Tits showed that if G is a Zariski-connected algebraic group defined over ℝ, then either G ⁢ ( R ) = G ⁢ ( R ) * or G ⁢ ( R ) / G ⁢ ( R ) * is a direct product of cyclic groups of order two (see [4, Theorem 14.4]). Hence, the result follows immediately from the above. ∎

Remark 3.3

Let 𝐺 be an algebraic group over ℂ, which is not necessarily Zariski-connected. Let G ⁢ ( C ) denote the complex point of 𝐺. Let k ∈ N . Then it is immediate that P k ⁢ ( G ⁢ ( C ) ) is dense in G ⁢ ( C ) if and only if 𝑘 is co-prime to the order of G / G 0 .

Corollary 3.4

Let 𝐺 be as in Theorem 1.1. Let G ⁢ ( R ) = L ⁢ ( R ) ⋉ R u ⁢ ( G ) ⁢ ( R ) and k ∈ N . Then P k ⁢ ( G ⁢ ( R ) ) is dense if and only if P k ⁢ ( L ⁢ ( R ) ) is dense.

Proof

The proof follows using the same procedure as in the proof of Theorem 1.1 applied to the group A = G ⁢ ( R ) . ∎

Remark 4.1

In view of (1) ⇒ (2) in Theorem 1.1, we observe that P k ⁢ ( A ) is dense implies 𝑘 is co-prime to the order of A / G ⁢ ( R ) * and P k ⁢ ( G ⁢ ( R ) * ) is dense. By [1, Corollary 1.3], it follows that P k ⁢ ( A ) is dense for all 𝑘 if and only if 𝐴 is weakly exponential.

Remark 4.2

Suppose 𝐺 is a Zariski-connected complex algebraic group defined over ℝ. Then the following are equivalent: (i) the map P 2 : S ⁢ ( G ⁢ ( R ) ) → S ⁢ ( G ⁢ ( R ) ) is surjective, (ii) the image of the map P 2 : G ⁢ ( R ) → G ⁢ ( R ) is dense, (iii) G ⁢ ( R ) is weakly exponential. This can be seen from [7, Theorem 1.6] and the fact that G ⁢ ( R ) / G ⁢ ( R ) * is a group of order 2 m for some 𝑚. It can also be deduced from Theorem 1.1. Indeed, by [7, Theorem 5.5] and Lemma 2.5, statement (i) implies 2 is co-prime to the order of the group G ⁢ ( R ) / G ⁢ ( R ) * and P 2 ⁢ ( G ⁢ ( R ) * ) is dense. Thus, by Theorem 1.1, P 2 ⁢ ( G ⁢ ( R ) ) is dense. Now (ii) ⇔ (iii) follows from Remark 4.1.

Proposition 4.3

Let 𝐺 be a connected linear Lie group. Then 𝐺 is weakly exponential if and only if P 2 ⁢ ( G ) is dense.

Proof

We assume that P 2 ⁢ ( G ) is dense in 𝐺. Since 𝐺 is linear, so is G / Rad ⁢ ( G ) (see [19, Proposition 5.2, p. 26]). Now, as G / Rad ⁢ ( G ) is a connected linear semisimple Lie group, it is isomorphic to A ⁢ ( R ) * for some Zariski-connected (semisimple) algebraic group 𝐴 defined over ℝ. Since P 2 ⁢ ( G ) is dense in 𝐺, it follows that P 2 ⁢ ( A ⁢ ( R ) * ) is dense in A ⁢ ( R ) * . Using Theorem 1.1, we see that

is surjective. Now apply [7, Theorem1.6] to see that A ⁢ ( R ) * is weakly exponential. Thus G / Rad ⁢ ( G ) is weakly exponential. As Rad ⁢ ( G ) is connected solvable, it is weakly exponential, and hence, by [16, Lemma 3.5], 𝐺 is weakly exponential. The other part is obvious. ∎

Corollary 4.4

Let 𝐺 be a Zariski-connected complex algebraic group defined over ℝ. Let 𝑁 be a Zariski-connected, Zariski-closed algebraic normal subgroup of 𝐺 defined over ℝ. If both P k ⁢ ( G ⁢ ( R ) / N ⁢ ( R ) ) and P k ⁢ ( N ⁢ ( R ) ) are dense, then P k ⁢ ( G ⁢ ( R ) ) is dense.

Proof

As 𝐺 is Zariski-connected for any odd k ∈ N ,

are dense by Theorem 1.1.

Suppose both the images of

are dense. By Remark 4.2, G ⁢ ( R ) / N ⁢ ( R ) and N ⁢ ( R ) are weakly exponential and hence connected. Recall that, for any closed subgroup 𝐿 of a topological group 𝐻, 𝐻 is connected if H / L and 𝐿 are connected. This implies G ⁢ ( R ) = G ⁢ ( R ) * . Since both the groups G ⁢ ( R ) * / N ⁢ ( R ) * and N ⁢ ( R ) * are weakly exponential, by [16, Corollary 2.1A], we have G ⁢ ( R ) * is weakly exponential. Therefore, P 2 ⁢ ( G ⁢ ( R ) ) is dense by Remark 4.2. ∎

Remark 4.5

Suppose 𝐺 is a Zariski-connected complex algebraic group defined over ℝ and k ∈ N . Then P k is surjective on G ⁢ ( R ) / G ⁢ ( R ) * and

together imply P k ⁢ ( Reg ⁢ ( G ⁢ ( R ) ) ) ⊃ Reg ⁢ ( G ⁢ ( R ) ) .

Remark 4.6

For a Zariski-connected complex algebraic group 𝐺 defined over ℝ, density of the image P k ⁢ ( G ⁢ ( R ) ) is equivalent to the fact that the complement of the image is of measure zero. This follows since density of the image P k ⁢ ( G ⁢ ( R ) ) implies Reg ⁢ ( G ) ∩ G ⁢ ( R ) ⊂ P k ⁢ ( G ⁢ ( R ) ) ( Reg ⁢ ( G ) is non-empty Zariski-open in 𝐺).

Proof of Corollary 1.2

By [7, Lemma 5.6], P k : G ⁢ ( R ) → G ⁢ ( R ) is surjective if and only if, for every unipotent element u ∈ G ⁢ ( R ) * , the map

is surjective. Note that Z G ⁢ ( R ) ⁢ ( u ) = Z G ⁢ ( u ) ⁢ ( R ) . Let H u = Z G ⁢ ( u ) for each unipotent element u ∈ G ⁢ ( R ) * . Since P k is surjective on S ⁢ ( H u ⁢ ( R ) ) , by [7, Theorem 5.5],

are surjective. By Lemma 2.5, P k ⁢ ( H u ⁢ ( R ) * ) is dense in H u ⁢ ( R ) * , which implies P k ⁢ ( H u ⁢ ( R ) ) is dense in H u ⁢ ( R ) by Theorem 1.1. This completes the proof. ∎

Corollary 5.1

Let 𝐺 be a connected real algebraic group. Then G ⁢ ( R ) * is exponential if and only if Z G ⁢ ( R ) * ⁢ ( u ) is weakly exponential for all unipotent elements u ∈ G ⁢ ( R ) * .

Proof

By McCrudden’s criterion, G ⁢ ( R ) * is exponential if and only if

is surjective for all k ∈ N . Further, by Corollary 1.2, P k : G ⁢ ( R ) * → G ⁢ ( R ) * is surjective if and only if P k ⁢ ( Z G ⁢ ( R ) * ⁢ ( u ) ) is dense in Z G ⁢ ( R ) * ⁢ ( u ) for all unipotent elements u ∈ G ⁢ ( R ) * . Now, by Remark 4.1, we conclude that Z G ⁢ ( R ) * ⁢ ( u ) is weakly exponential. ∎