R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
S URJIT S INGH
Pseudo-valuation and pseudo norm
Rendiconti del Seminario Matematico della Università di Padova, tome 51 (1974), p. 83-96
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SUBJIT SINGH
(*)
1. Introduction.
P. M. Cohn in
[1]
hasgiven
necessary and sufficient conditions fora non-discrete
topological
field to have itstopology
inducedby
apseudo
valuation. For this purpose, he introduced the notion of gauge set. In thisnote, using techniques
somewhatanalogous
to those ofCohn,
westudy pseudo
normedtopological
vector spaces overpseudo
valuated fields. We extend the
concept
of gauge set topair
gauge set and use bounded sets intopological
fields andtopological
vectorspaces over
topological
fields. We also need the notion oftopologically nilpotent
elements to achieve ourgoal.
The main aim of this note is to
give
necessary and sufficient con-dition for a
topological
space V on atopological
field K to have theirtopologies
inducedby
apseudo
norm on Vcompatible
with apseudo
valuation on .K. The
topology
on K is inducedby
thispseudo
valuation.We
give
all the relevant definitions andproperties
to make thenote self contained.
2. Pseudo valuation and
pseudo
normed vector spaces.In this section we shall introduce the basic notions and
give
ex-amples.
(*)
Indirizzo dell’A.:Dept.
of Mathematics,Panjab University,
Chan-digarh -
14, India.The Author wishes to thank Dr. N. Sankaran for his constant encour-
agement and advice and to
acknowledge
theUniversity
Grants Commission for financial support.DEFINITION 1. Let K be a field and R be the field of real numbers.
A
mapping
w from .K tosatisfying
thefollowing properties
is calleda
pseudo
valuation.From this definition it is easy to observe the
following:
EXAMPLE 1. Let .K be any field. Set
Then it is
easily
seen that c~o is apseudo
valuation.EXAMPLE 2. Let .I~ _ .1~ be the field of real numbers. If we set
OJ1(a)
== wherela I
is theordinary
absolute value of a then is apseudo
valuation.Setting w2(a)
=[
where c is areal number
larger
than1,
we find W2 is also apseudo
valuation.DEFINITION 2. Let
(K, co)
be a field with apseudo
valuation and V be a vector space over ~.By
aco-pseudo
norm¡I...II ))
on TT we meana real valued function on TT such that
(i) 11 v )) > 0
for all v in V andII ==
0 where 0 is the zero vector ofV;
(ii) ))
for all a in .g and v inV;
(iii) [[v~ -- v2 J[ ~ [J2’1 [[ -~- II V2 11
for all vI, v2 inV;
(iv)
there exists v in V such thatWe shall hereafter refer the
cv-pseudo
norm asjust pseudo
norm.We observe that
11- v 11 = 11 v II
for all v in TT and + v211))
EXAMPLE 3. The field R with the
ordinary
absolutevalue,
is afield with
pseudo
valuation. Let V be the n dimensional vector spaceover R.
Define I
Then11...11 11
is apseudo
norm on V.EXAMPLE 4. Let
(K, roo)
where roo is thepseudo
valuation definedin
example 1,
be a field with apseudo
valutation.If we take for V the one dimensional vector space over K itself then
setting 11 v ~~ = wo( v),
weget
apseudo
normed vector space.Clearly
apseudo
norm on a vector space defines atopology
on Vand
pseudo
valuation defines atopology
on the field in such a way that V is atopological
vector space and K is atopological
field. Theneighbourhood
bases of zero are defined as follows.For
(.K, OJ),
the collection U= (U§)
whereand for
( V, II...11 ~I ~,
the collection wheregive
theneighbourhood
basis. There is aunique topology
íw onK,
and -r 11 11 on TT with
respect
to which U and 9Z are basis ofneighbour-
hoods at a and v
respectively.
It is not hard toverifiy
that under thesetopologies
.K becomes atopological
field and TT atopological
‘
vector space.
A
pseudo
valuation co(and
apseudo
norm[[... [~
on a field .K(on
avector space
V)
is said to be trivial if thetopology
on .K(on V)
isdiscrete.
Examples
1 and 2give
trivialpseudo
valuation on K whileexample
4 illustrates a trivialpseudo
norm on a vector space.Next,
if
(V, 11. 11) )
is apseudo
normed vector space over a non-trivialpseudo
valuated field
(.K, a~),
then thetopology
inducedby 11.11 ))
on Tj is non-discrete.
DEFINITION 3. A subset A in a
topological
fieldr)
is calledbounded,
ifgiven
anyneighbourhood Ul
of0,
there exists aneigh-
bourhood
U2
of 0 such that HereAnalogously,
we define a subset .B of atopological
vector space(V, T)
over atopological
field(K, r),
ifgiven
aneighbour-
hood N of 0 in V there can be found a
neighbourhood
U of 0 in K such that1. The set K in the
topological
field(.~, i)
can not be bounded unless -r is discrete.Likewise,
the set V of thepseudo
normedvector space
( V, T)
on(.g, i)
can not be bounded unless thetopology
7:on .g is discrete.
LEMMA 1.
If BI
andB2
are two bounded subsetsof ( V, í2)
over(K, i1)
then
.B1-f- B2
and aBI for
all a E K are also bounded.PROOF. Since
.B1
andB2
arebounded,
there exist suitableneigh-
bourhoods
Le1
andU2
of 0 in .K such thatUi - Bi c
N’ for anyneigh-
bourhood N’ of 6 in ~’. If N is any
given neighbourhood
of 6 inV,
then
choosing
.N’ to be such thatN’ -~-
N’ c N and U to beU,
(1U2 ,
we find that
From the
continuity
of scalarmultiplication,
weget
for anygiven neighbourhood NI
of0,
anotherneighbourhood N2
of 6 andU,
of asuch that As
BI
isbounded,
there exists aU2
such thatU2.BIcN2.
ThereforeThus a - B, is
bounded.COROLLARY 1.
I f Al
andA2
are bounded subsetsof (K, i)
then soare
A 1 ; A2
anda.Al f or
all aEK.COROLLARY 9.
I f
B is a bounded subseto f ( TT, í 2)
over(.K, zl)
then~v BI
where v i1~V,
is also a bounded subsetof
V.The
proof
ofCorollary
1 follows fromspecializing
T~ to I~ in lemma 1and of
Corollary
2 from the fact thatsingleton
sets arealways
bounded.COROLLARY 3.
If
A is a bounded subsetof (K, -r)
and a is any ele- mentof K,
thenla + Al
is also bounded.LEJDIA 2.
Suppose
that(K, m)
is afield
withpseudo
valuation and
( V, 11. ~l ~
is apseudo
normed vector space K. Then(i)
.A. is exists a real number r such thatfor
all cx in A.(ii)
is exists a real numbC14 s such that||v||s for all v in B
PROOF. It is sufficient to prove
(ii)
as(i)
can be deducedby
tak-ing
V = .K and11. II
= cv. If B isbounded,
then for anygiven 8
>0,
we can find a ð > 0 such that if
w(a) ð,
for some c~ E.K,
thenIla -
vII
8for all v in B.
Taking e
=1 and acorresponding ð,
from the non-triviality
of (J) we can find at least one non-zero a in .g with 6such that
~~ v ;~ = ~~ w(a-1)
for all v in B. Ifwe take s =
w(a-1),
weget
the desired result.Consersely,
suppose there exists an s with for all v in B.Then, given
any e >0,
if we take 6 _~/(1-~-- s)
then whenever3
for some in H then
for all v in B. Thus
3. Pair gauge sets and
pair
gauge functions.P. M. Cohn introduced in
[1]
theconcepts
of gauge set and gauge function tostudy pseudo
valuations on a field ..K. Here we extend these notions tostudy pseudo
normed vector spaces overpseudo
valuated fields.
DEFINITION 4. Let V be a v ector space over a field .K. An ordered
pair (H, G)
of sets where H c V and G c .~ will be called apair
gauge set if thefollowing
five conditions hold.(i)
(ii)
and where andG-G =
(iii)
There exists a c in _K with andHere .H~ -}- H = {v
andG -f-
G =~c~
+(iv) H(G)
is a proper subset ofV(K).
(v)
There exists a non zero element d in G such that VAny
element dsatisfying
condition(v)
of the above definition will be called a gauge element of thepair
gauge set(H, G).
EXAMPLE 5. Let
(.K, ÚJ)
be a non-trivialpseudo
valuated field with and(V, [; . ~[ )
be apseudo
normed vector space over(I~, cc~) .
If we take H - and G = then(H, G) gives
rise to apair
gauge set and any non-zero element d in G with 1 is a gauge element.LEMMA
3. If (H, G)
is apaii-
gauge setof ’a
vecto’f space V over K and d is a gauge elementof (H, G),
thenare
strictly
sequences andPROOF. we
only
prove(i).
From d in G andH,
we derivethat dn H = dl-1 -
(d WI ) c
for all n. Strictinequality
results fromthe fact that and H is a proper subset of V.
REMARK 2. If d is a gauge element of a
pair
gauge set(H, G)
thenso is
dt(t > 0).
We prove it thus:implies
that thereexists an
integer t >
0 such that c c d-t G. Hence G + G C c ~{d--t ~
.Ghd-t.G and
Therefore,
when the gauge element is not fixed inadvance,
we maychoose d such that and
With each
pair
gauge set(~, G)
and gauge elementd,
we associatea
pair
of extendedinteger
valued functions g andf
on V and K respec-tively,
as follows:(1)
If and I theng(v)
= oo:if and then
99(v):= n.
(2) Suppose
and Then we set ifand We also set
f(0 ) =
oo.(g, f )
is calledpair
gaugefunction
associated with thepair
gauge set( H, G)
and gauge element d.Let 1T be a vector space over .K and
(~I, G)
be apair
gauge set with d as a gauge element. If(g, f )
is thepair
gauge function associated with(.H, G)
andc~,
then we have(i) 99 (v)
=g(- v)
for all v E V andf (-1 )
= 0 .(ii) --E- cp(v)
for all a in .K and v EV,
andf(a-b)>
~ f (c~)
where a and b vary over K.(iii)
There exists apositive integer t
such thatfor all
and
Given any
pair
of extendedinteger
valued functionsf )
on Vand K
satisfying
the aboveproperties
with some non zero d inK,
we
get
apair
of sets H and Ggiven by
H =(v «
and G =- ~aE.K~f(a)~0~ .
One can see without
difficulty
that(H, G)
is apair
gauge set with d as the gauge element and(q, f )
is thepair
gauge function associated with( H, G )
and d.REMARK 3. If t is as in
(iii)
above and we take dt as gauge element instead ofd,
then thepair
gauge functionsf’)
associated with(H, G)
and d’satisfy
theinequalities
of(iii)
with treplaced by 1,
Thatis to say +
V2)
-1 and4. The
pseudo
valuation andpseudo
norm definedby
apair
gauge set.Let V be a vector space over a field
.~, (H, G)
be anypair
gauge set with gauge element d and(g, f )
thecorresponding
gauge function.By
remark 3 above we may choose a d so that t = 1.(Here t
is asin the
properties
of thepair
gaugefunction).
Let now e be a real numbersuch that
1 c e c 2.
We setand
Obviously w and g
are real valued functions on V and Krespecti-
vely. Moreover g
satisfies the condition of apseudo
valuation withg(a + b) e. g(b)}
for alla, b
in ~.And 1p
fulfills the con-dition of a
pseudo
norm with4. If
v,, V2 in V are that and +V2)
>>
1p(Vl)
+1p(V2)
thenc ~(v2)
where e and y~ are as above.PROOF.
Suppose 1p(v2).
ThenThis contradicts the
hypothesis.
Hence~(v~).
In case =0,
then = 0
~(v~).
Ifnot,
then from the definitionof 1jJ,
weget q;(v2)
so that 1 +q;(v2)
as q; is aninteger
valued func- tion.Consequently,
LEMMA 5.
Suppose
v = vl -f- + ... + Vn is adecomposition of
v V.Then we can
find
anotherdecomposition of
such
that’
I and1p(v)
where e are asabove.
PROOF. In case + ... -r- then take v = WI and 0 for i =
2, 3,
... , m. Then the result follows.Otherwise,
we claimthat there is a
decomposition
of v as v = wl+
W2 --f-- ... + wrn such thatIf itself has the
property (A )
wekeep
it.Otherwise,
theresuch that Then
In this
decomposition
If thisdecompo-
sition of v as a sum of
v’
satisfies(A)
we are done. In thecontrary
case we shall continue with this process till we
get
the desiredtype
ofdecomposition.
Note that m ~ 2.Arrange
the wj such thaty(wl)
c ~(w2) ~ ... c y~(wm).
Thenby
lemma3, for j
=1, 2,
... , m -1. We now claim that + ... + for all t = 1
2,
..., m. This followsby
induction and weget V(v)
==e-w(wm).
THEOREM 1. Let V be a vector space over
K, (H, G) pair
gauge seton V and
g)
be two real valuedfunctions defined
as above.I f
we setand ’
where the
infima
are taken over allpossible decompositions o f
a=== ~
a,and v
=== 2
Vj then w is apseudo
valuation on K11...11 11
is apseudo
norm on V
sa,tisfying f or
all a inK,
andPROOF. We prove
only
thepseudo
normpart.
It is clear that for all v inV,
For the otherinequality,
take anydecomposi-
tion of v as v = VI + ... + v, and
apply
lemma5,
toget
another de-composition
of v as v = w1-E- ... -~- w. such that andlower bounds over all
possible decompositions
of v weget ~~~) c e ~ 11 v 11.
We now
verify
that11 - - - 11
is aw-pseudo
norm on V. That and~~ 8 (~ = 0
are obvious from the definition.Since V
is notidentically
zero on
V,
we see from theinequality ~(v) c e ~ ))v II
and the fact that e >1,
that11 v 11
is notidentically
zero. To show that11 v 11
for all a in K and v in
V,
we take for any e >0,
adecomposition
ofa = a2 -T- ...
+ an ,
a, and adecomposition
of v = v1-~- ... -+- vm withVi E V
such thatand
Then
In a similar fashion we can show that
5. The
topology
associated withpair
gauge sets.Given a
pair
gauge set(H, G)
of a vector space V over a field Kwe can associate a
pseudo
valuation co on K and apseudo
normII...11 11
on V. Note that OJ andII...11 11
stilldepend
upon the choice of the gauge element d. We will now show that m andII...11 ))
areindependent
of d
upto equivalence (*).
THEOREM 2. Let
(H, G)
be apair
gauge setof
a vector space TT over K.Suppose l;
and co arerespectively
thepseudo
norm andpseudo
valua-tion on V and K.
If d is
any gaugeof (H, G),
then and~dn ° G} as n
varies on the naturalnumbers, form
a basisof neighbour-
hoods
of
0 in TT ando f
0 in K in thetopology defi1ted by 11... 11
and a).PROOF. The
proof
thatfd,GI
forms aneighbourhood
basis of 0in H in the
topology
definedby
a) may be found in[1].
For the vectorspace
part
weproceed
thus. Set U ={v
c(( C
where 1 C e 2.It is easy to observe that U is a
neighbourhood
of 0 in thetopo-
logy
definedby 11 ... 11.
Sincey(v)
ellvll for
all v inV,
ifwetakeuc U,
then This means 11, E H. Thus H itself is a
neigh-
bourhood of 0 in thetopology
definedby II...B1. Continuity
of scalarmultiplication
shows that for all non-zero a in Kgive neigh-
bourhoods of 0.
Therefore,
for any choice of gauge element d of(G, H), give neighbourhoods
of 0 in thetopology
definedby 11 ... 11
on V.Let now
dl
be another gauge element of(H, G)
used todefine w
on V.Then
dl’
c for some t ~ 0 so that dt E If N is anyneigh-
(*)
Twopseudo
valuations on a field .K(pseudo
norms on a vector spaceV)
are said to be
equivalent
ifthey
define the sametopology
on K(V).bourhood of 0 in the
topology
definedby II...11 11 on V,
then there existsa real number e > 0 such that for all elements with
11 v ||
s, v E N.Choose a
positive integer n
such that e-n c. If v E thenc y~ (v ) c e-~2 C E .
This shows v E N. Thus and henceThus
gives
a basis ofneighbourhoods
at 0 of V withrespect
to thetopology
definedby 11 ... 11.
6. Characterization of
pseudo
normtopology
on a vector space.In this
concluding
section wegive
a necessary and sufficient con-dition for a
topological
vector space,(V, T2)
over atopological
field(K, Z’1)
to have theirtopologics
inducedby
apseudo
norm11...11 ))
on Vand a
pseudo
valuation oi on .g. We need the notion of atopologically nilpotent
element in atopological
field.DEFINTION 5. An element ac of a
topological
field H is said to betopologically nilpotent
if an -* 0 in thetopology
of K.THEOREM 3..Let
(K,
be a non-discretetopological field
and(V, ~’2)
be a
topological
vector space over(K, T1).
ThenTl
can be obtainedby
a
pseudo
valuation w on K rxndT2 by
a non-trivialo)-pseudo
norm
II...11 11
an Vif
andif
(i)
Knon-empty
open bounded set.(ii)
There existg ac non-zero,topologically nilpotent
element in K.(iii)
TT contains anon-empty
open bounded set.PROOF.
Suppose T1
is definedby
apseudo
valuation a) on K andT2 by
aw-pseudo
normII...11 11
on V. Then if we setand
then it can be seen without
difficulty
that A and B are both non-empty
open sets. Lemma 2 enables us to check that sets A and Bare bounded in .K and V
respectively.
From thenon-triviality
of w,we can find a non-zero element a in g such that
w(a)
1 and th is acis a
topologically nilpotent
element.7
For
establishing
the converse, weadopt
thefollowing strategy.
We construct from the
given
conditions apair
gauge set and show that thepseudo
norm andpseudo
valuation associated with thispair
gauge set
give
rise to thetopology Tl
on 1~ andT2
on V.Let .~. be a
non-empty
bounded open set in H. Then as in[1~,
weconstruct a set G and an element
d ~
0 in G which istopologically nilpotent
such that is aneighbourhood
basis of 0 in H withrespect
toTI
andsatisfy
the conditions of a gaugeset, namely, -1
EG,
w
G ~ G c G ; G cK;
for someinteger ~0;
1
Next we shall construct H. Let B be the
given
open bounded set in V. Then without loss ofgenerality
we may assume that B is a sym- metric open bounded setcontaining
0. Thus B is aneighbourhood
of 0. The collection
~b ~ B~
as b varies over the non-zero elements of Kgives
rise to aneighbourhood
base at 0.For,
if N is anyneighbourhood
of
0,
from the boundedness ofB,
we can find aneighbourhood
U of 0in ~K such that U ~ B c N. As
TI
is notdiscrete,
we can find a b cU, b ~ 4
such that b. BeN.Using continuity
of scalarmultiplication
and the fact that B is aneighbourhood
of 6 we find that(ds - G).
c B for some non nega- tiveinteger s
and a b’ inK, b:A 0.
As is aneighbourhood
of 0Hence we can find an
integer
such thatC b’ ~ ds- G. Therefore d t ~ G ~ B C B.
Now set dt B. Then
clearly A1
is asymmetric,
bounded open setcontaining
0.Consequently,
as b varies over the non-zeroelements of .~ forms a
neighbourhood
basis of 0.Finally
define This H will serve our pur- pose. It iseasily
observed that H is an open,bounded, symmetric neighbourhood
of 0. We assert that forms a basis ofneighbour-
hoods at 0.
For,
if N is anyneighbourhood
of0,
then we can find anon-zero b in K such that As
dn -~ 0,
dn. b-1 also tends tozero.
Therefore,
dn. b-1belongs
to dt- G for someintger
t. Thus dn E b·~ dt~ G. Hence dn~ H C
(b.
dt~G) H ~ b(dt. GH)
= N.We now show that
(H, G)
is apair
gauge set with d as the gauge element.Symmetry
of Hgives - H
=H,
and from the definition ofH,
we havealready
noted that -1 E G andFrom the
continuity
of addition in V and the fact thatH}
is a basis of
neighbourhood
of0,
weget
for some in-teger p ~ 0.
Already
we have seen thatG ~- Taking
c = weget
V since g is bounded while V is not. as
noted above. For any v e
V, dnv
- 0 as n - co since d istopologically nilpotent. Therefore,
for someinteger
~2 ~ 1. ThusHence and K is contained in Thus
(G, H)
isa
pair
gauge set. The rest of theproof
follows from theorem 2.We conclude this note with an
example
of atopological
vectorspace
(V, i)
over apseudo
valuated field(K, co)
such that T is notgiven by
any(o-pseudo
norm.EXAMPLE 6. Let .~ = 1~ be the field of real numbers and
v(a)
-= c -
| a I where I
is theordinary
absolute value of a and c is a real number such that1 ~
c 2.Clearly
a) induces apseudo
valuation on K.For the vector space V we take the
ring
ofpolynomials
in one variable xover K and
give
atopology
T for V as follows.Set Q = =
i1.)
sequence ofpositive
realnumbers}.
Foreach d in
Q,
define ,It can be verified without
difficulty
that the set 91 =gives
riseto a
neighbourhood
basis for 0 in V. If T is thetopology given by in
then
(TT, T)
is atopological
vector space over(.~, w).
We now show thatgiven
anyNiX
in9~
we can find aNfJ
such that for all non-zero ain K, a ~ Na ~ N~ .
Forthis,
if o~ = CX2,... )
thentake fl = ({3I, ... ) where Bn
= 27i an for each n.Suppose
b is a non-zero element of K.Then there exists an
integer
such thato)(b)
> Thepolynomial (fJm/2)
xmbelongs
toN f3,
while b -f3rn/2’
x does notbelong
toNiX’
For,
Thus From this one can
easily
infer that there does not exist anon-empty
open bounded subset of V.Therefore,
in viewof Theorem
3,
thetopology
r for TT can not be inducedby
a(o-pseudo
norm.
REFERENCES
[1] P. M. COHN, An invariant characterization
of pseudo
valuation on afield, Cambridge
Phil. Soc., 50 (1954), pp. 159-177.Manoscritto pervenuto in redazione il 23