Homomorphs of Knot Groups
F. Gonzalez-Acuna
The Annals of Mathematics, 2nd Ser., Vol. 102, No. 3. (Nov., 1975), pp. 373-377.
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Annals of Mathematics, 102 (1975), 373-377
Homomorphs of knot groups
Neuwirth's problem U ([8], [9]) asks whether a finitely generated group
with deficiency 5 0 and weight one is a homomorph of a knot group. We will
show here t h a t the answer is affirmative even if the condition on the deficiency
is omitted.
THEOREM
1. A group i s a homomorph of a kxot group i f and only i f i t
i s jinitely generated and has weight at most one.
The idea of the proof is to realize a finitely presented group of weight
one as the fundamental group of the complement of a 2-sphere S2in a manifold M 4of the form S2x S2#
# S2x S2and then to find a 3-sphere S3in
M 4intersecting S2along a knot k such t h a t the inclusion induced homomorphism n1(S3- k) n1(M4- S 2 )is onto.
I would like to thank Dennis Roseman for stimulating conversations. I
a m grateful to Professor J. Milnor for pointing out t h a t our result, originally
stated for finitely presented groups, actually holds for finitely generated
groups.
a
a
a
-
1. Main results
We work in the smooth category. Homomorphisms between fundamental
groups of spaces are induced by inclusion maps unless otherwise stated. The
connected sum of r copies of a manifold M will be denoted by r M .
A knot group is the fundamental group of the complement of a 1-sphere
in S3.
The weight of a group G is the smallest number w such t h a t there exist
w elements g,,
., g , in G with the property t h a t the normal closure of
{gl, ..., g,} is G. Clearly a homomorphic image of a group of weight a t most
w has weight a t most w.
Suppose K w 7c1(S3- k) where k is a 1-sphere in S3. Let a,, . ., a, E K.
If we spin ([I]) the knot k, we obtain a 2-sphere S2in S4with 7c1(S4
- S2)w
K. Now find disjoint embeddings 9,: S' x D 3 S4- S" i = 1, .. r which
represent, up to conjugacy, the elements a,, i = 1, ..., r . By performing
surgery on S4using the embeddings 9, we obtain a $-manifold M 4 and a 2-+
a ,
sphere S h m b e d d e d in it, with n,(M4 - S 2 )w K/(a,, .. ., a,) where ( )
denotes normal closure. If one takes the embeddings g,in such a way t h a t
M 4is stably parallelizable ([6, Lemma 5.4, p. 5141) then M 4will be diffeomorphic to r ( S 2 x S2).
If a finitely presented group G is a homomorphic image of a knot group
K then there are elements a,, ..., a, E K such t h a t G w K/(a,, ..., a,) ([2,
Proposition, p. 4941). By the above construction G is then the fundamental
group of the complement of a 2-sphere embedded in r ( S 2 x S2).
Motivated by these observations we state the following:
THEOREM
2. Let G be a finitely presented group. Then the following
conditions a r e equivalent:
(i) G i s a homomorph of the group of a k-complement link i n S3.
(ii) G has weight a t most k.
(iii) There i s a p a i r (M4,L2)with n1(M4- L2) w G where M4 = r ( S 2 x
S 2 )for some r and L 2i s the union of k disjointly embedded 2-spheres.
-
Proof. We will show t h a t (i) (ii) -- (iii) -- (i).
( i ) (ii). An easy application of van Kampen's theorem ([Ill)shows t h a t
the group of a k-component link in S3has weight k so t h a t , if G is a homomorph of such a group, G has weight a t most k.
-
(ii) (iii). Suppose the weight of G = j x,,
x,: r,, . ., T , j is not
greater than k. Then there are words r,,,,
r,,, such t h a t I x,, ., x,:
r,,
r,,, I is the trivial group. Let N,4 = n(S1 x S3). Identify n,(N,4)with
the free group having x,, ..., x, a s basis and represent the elements r,, ...,
r,+, by disjoint embeddings q,: S' x D 3 N,4, i = 1, .. -,p + k.
We may assume the q, are so chosen t h a t the manifold N,4 = ~ ( q , , . ,
qp+,), obtained by performing surgery, is stably parallelizable ([6, Lemma
5.4, p. 5141). Then N,' is a simply connected manifold with an even quadratic
form of index 0 so that, by [7, Cor. 3, p. 1261, N,"has the homotopy type of
a connected sum of copies of S h S2.
We have disjoint embeddings 9:: D 9 S2 N,", i = 1, .. ., p + k such
t h a t the complement in N,4 of U:=lybti(Int D 2 x S2) is obtained from the
disjoint union
a ,
a ,
a ,
-
-
(N,"-
u:
y,(S1 x Int D3))U ( D 2 x S2), U ( D 2 x S2)2U
.
U ( D 2 x S2),
by identifying yi(u, v), for u E S', v E S 2 ,with (u, v) E (D2x S2),,i = 1, .. p.
By the theorem of van Kampen we have n,(N," - L" w G where L"
UL,yb+((O x S2). By [12, Theorems 2 and 31, if we take the connected sum
of N,' with m(S2x S2)along a kdisk disjoint from L 2we obtain, if m is large,
a ,
375
HOMOMORPHS O F KNOT GROUPS
-
a manifold M 4 diffeomorphic to r ( S 2 x S 2 ) for some r. L 2 is naturally
embedded in M 4and, since m ( S h S 2 )is simply connected, n1(M4- L"
G.
(iii) -- (i). The manifold M 4can be expressed a s M 4 = DD" U ( H 3 D2) U
D: where H V s the union of 2 r disjoint 2-disks,
and
We can assume t h a t L 2 n D = 0 and t h a t L 2n ( H z x D2) = F x D 2
where F is a finite subset of Hz.
Let Mi = M 4 - Int D L Then M,' - L2can be expressed a s the union of
D 5 - L 2n D: and ( H z - F ) x D2. Using van Kampen's theorem one sees
nl(M,4 - L2)is onto and nl(M,4 - L2) n1(M4- L2)
t h a t n,(D+ L2 n D:)
n1(M4- L2)is a n epimorphism.
is an isomorphism so t h a t nl(D> L2n D:)
If 0 is the center of D h e may identify D4 - {0} with S3x (0, 11 and
we have a projection map p from D: - {0}onto (0, 11. By slightly altering
L 2 n D h e find a properly embedded 2-manifold S in D: - {0} such t h a t
(D:, S) is homeomorphic to (04, L 2n 0 4 ) and f = p I S is a Morse function
with a s many critical values a s critical points. S consists of k components
of genus zero.
Following [4] we say t h a t a critical point c of f is a positive (resp. negative) saddle point if c is of index 1and, for sufficiently small E > 0, the number
of components of f -'(f(c) + E) is greater (resp. smaller) than the number of
components of f -'( f (c) - E).
We may assume t h a t f (c) < f (c') if the index of the critical point c is
smaller than the index of the critical point c' or if c is a negative, and c' a
positive, saddle point (see, for example, [lo, Lemma 3, p. 4351).
Let a. E (0, 1) be a regular value of f such t h a t f (c) < a if c is a critical
point of index 0 or c is a negative saddle point and f (c) > a if c is a critical
point of index 2 or c is a positive saddle point. Then f-'(a) is a k-component
link in p-'(a) = S3x {a}.
A presentation of n,(D?S ) can be obtained by adjoining relations to
a presentation of n,(p-'(a) - f-'(a)), one for each critical point of index 1
([3, p. 1331). It follows t h a t n 1 ( D 5 L 2n D:), and therefore n,(M4- L2),is
a homomorph of the fundamental group of the complement in S3x {a}of the
k-component link f-'(a). This completes the proof of the theorem.
-
-
+
J. Milnor has pointed out t h a t a finitely generated group G= I x,, .. ., x,:
r,, r,, .. / of weight w is a homomorph of a finitely presented group of
weight w. In fact, if y,, . ., y, a r e words such t h a t I x,, . x,: y,, ., y,,
r,, r,, . I is trivial, then one can find s,, . ., s, in the normal closure of
, ' ) . finitely presented
{r,, r,, .}, where si has the form ~ , I I , ( u , ~ y ~ ~ ~ uThe
group 1 xl, .., x,: sl, . S, I of weight w has G as a homomorph.
Together with Theorem 2 this yields the following result, which has
Theorem 1 a s a special case.
a ,
a ,
THEOREM
3. A group i s a homomorph of the group of a k-component
link i n S3if a n d only if i t i s finitely generated and has weight a t most k.
COROLLARY
1. A finitely generated simple group i s a homomorph of a
knot group.
Proof. A nontrivial simple group is the normal closure of any of its
nontrivial elements and, therefore, has weight one.
2. A solvable group i s a homomorph of the group of a kCOROLLARY
component link i n S3if a n d only if i t i s finitely generated a n d i t s abelianixation can be generated by k elements.
Proof. It suffices to show t h a t a solvable group G has weight a t most k
if GIG' can be generated by k elements. If g,, . , g, a r e elements of G
whose images in GIG' generate GIG', then G/(g,,
g,) is perfect, solvable
and, therefore, trivial; hence G has weight a t most k.
a ,
2. Finite homomorphs
In the form given in [8], problem U asked whether a finitely presented
group whose abelianization is cyclic is a homomorph of a knot group. Even
though, for infinite groups, the property of having cyclic abelianization is
weaker than t h a t of having weight a t most one ( [ 5 , p. 117]), it seems reasonable to expect t h a t , for finite groups, the question in this form has a n
affirmative answer. One may therefore ask the following:
Question. Does every finite group with cyclic abelianization have weight
a t most one?
P. Kutzko has answered this question in the affirmative ([13]). More
generally he shows t h a t , for a finite nontrivial group G, w(G) = max { I ,
w(G/G1)}where w denotes weight. It follows t h a t finite group G is a homomorph of the group of a link with k components, k 2 1, if and only if GIG'
can be generated by k elements.
HOMOMORPHS OF KNOT GROUPS
377
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(Received May 19, 1975)