Hongmei ZHU(朱紅梅)
College of Mathematics and Information Science,Henan Normal University,Xinxiang 453007,China
E-mail:zhm403@163.com
In Finsler geometry,Douglas curvature is an important projectively invariant,which is introduced by J.Douglas in 1927.It it also a non-Riemannian quantity,because all the Riemannian metrics have vanishing Douglas curvature inherently.Finsler metrics with vanishing Douglas curvature are called Douglas metrics.Roughly speaking,a Douglas metric is a Finsler metric which is locally projectively equivalent to a Riemannian metric[5].
Douglas metrics form a rich class of Finsler metrics including locally projectively flat Finsler metrics[4]and Berwald metrics;The later are those metrics which Berwald curvatures vanish[1].It is known that a Finsler metric is locally projectively flat,namely,its geodesics are all straight line segments in some suitable locally coordinate system,if and only if its both Douglas curvature and Weyl curvature vanish.
A Randers metric,introduced by a physicist G.Randers in 1941,is of the form F=α+β,whereis a Riemannian metric and β=bi(x)yiis a 1-form.However,it can also be expressed in the following navigation form
(α,β)-metrics are a rich and important class of Finsler metrics,partly because they are“computable”.The researches on(α,β)-metrics enrich Finsler geometry and the approaches offer references for further study.In 2009,B.Li,Y.Shen,and Z.Shen gave a characterization of Douglas(α,β)-metrics with dimension n ≥ 3[5].
A more general class named general(α,β)-metric was first introduced in[11]in the following form
where α is a Riemannian metric,β is a 1-form,b:= ‖βx‖α,and φ(b2,s)is a smooth function.If φ = φ(s)is independent of b2,then F=is an(α,β)-metric.If α =|y|, β = 〈x,y〉,thenis the so-called spherically symmetric Finsler metrics[9,13,15].Moreover,general(α,β)-metrics include part of Bryant’s metrics[2,11]and part of fourth root metrics[7].In[9],X.Mo,N.M.Solórzano,and K.Tenenblat determined all of spherically symmetric Douglas Finsler metrics.Under the condition that β is closed and conformal,the author determined all of Douglas general(α,β)-metric[14].
In this article,we only need the condition that β is not parallel with respect to α determines all of Douglas general(α,β)-metric on a manifold of dimension n > 2.The main result is given below.
Theorem 1.1Let F= αφ(b2,s),s=,be a regular general(α,β)-metric on an open subset U ? Rn(n ≥ 3).Suppose that β is not parallel with respect to α and F is not of Randers type.Then,F is a Douglas metric if and only if F lies in one of the following cases:
(1)φ=φ(b2,s)satisfies
where ki:=ki(b2),i=1,2,3,4 and β satisfies
where τ= τ(x)is a scalar function on U.
(2)φ=φ(b2,s)satisfies(1.1)and
Moreover,β satisfies
where k:=k(b2)/=1,μi:=μi(b2),i=1,2 andμ1+μ2b2=(k+1).In this case,dβ /=0.
(3)φ=φ(b2,s)satisfies(1.1)and
Moreover,β satisfies
Let us take a look at a special case;When φ1=0,takingthe case 1 of Theorem 1.1 is reduced to a result which characterizes Douglas(α,β)-metrics with dimension n ≥ 3(Theorem 1.1 in[5]).Notice that Theorem 1.1 contains some important singular,Finsler metrics,for example Finsler metrics of Kropina type.
X.Wang and B.Li studied Douglas general(α,β)-metrics in[10].Their result is just case 1 of Theorem 1.1.However,we find that case 2 and case 3 also exist recently.For example,for case 2,takeand k=3,then in this case,(1.1)and(1.3)are reduced to the followings:
By a direct computation,it is easy to verify that the following function
satisfies(1.8)and(1.9),where f(t)is an arbitrary function satisfying the following inequalities:
In this case,the corresponding general(α,β)-metric is a regular Finsler metric.
In the following,according to(1.10),for case 2,let us take a look at an example.
Example 1.1Let M be a manifold of dimension n > 2.TakeandIt is easy to verify that α and β satisfy
Moreover,dβ/=0.Choose f=(t+2)2in(1.10),then
where b:= ‖β‖α=Hence,the corresponding Finsler metrics
are Douglas metrics.
Similarly,we can find regular Finsler metrics in case 3.Mostly important,β is closed in the case of(α,β)-metrics;However,it is seen from Theorem 1.1 that β permits not closed in the case of general(α,β)-metrics.That is to say,our results express that Douglas general(α,β)-metrics are much more and richer than Douglas(α,β)-metrics.
For three cases of Theorem 1.1,it is easier to solve case 1 because it is close to the result of Douglas(α,β)-metrics.However,case 2 and case 3 do not occur in the case of Douglas(α,β)-metrics,especially β is not closed.It is a pleasure and surprised to us.The two cases are complicated.
In local coordinates,the geodesics of a Finsler metric F=F(x,y)are characterized by
where
The local function Gi=Gi(x,y)define a global vector fieldon TM{0},which is called the spray.For a Riemannian metric,the spray coefficients are determined by its Christoffel symbols as
Douglas metrics can be characterized by
It is known that Douglas metrics can be also characterized by the following equations[1]
By definition,a general(α,β)-metric is a Finsler metric expressed in the following form,
where φ(b2,s)is a positive smooth function defined on the domain|s|≤ b < bofor some positive number(maybe in finity)bo.Then,the function F= αφ(b2,s)is a Finsler metric for any Riemannian metric α and any 1-form β if and only if φ(b2,s)satisfies
when n≥3 or
when n=2[11].
where(aij):=(aij)?1and bi:=aijbj.It is easy to see that β is closed if and only if sij=0.
According to[11],the spray coefficients Giof a general(α,β)-metricare related to the spray coefficientsαGiof α and given by
where
In this section,we are going to show the necessary conditions for a general(α,β)-metric to be a Douglas metric.
Lemma 3.1Let F= αφ(b2,s),s=be a regular general(α,β)-metric on an open subset U ? Rn(n ≥ 3).Suppose that β is not parallel with respect to α and F is not of Randers type.If F is a Douglas metric,then F lies in one of the following cases:
(1)φ=φ(b2,s)satisfies
where ki:=ki(b2),i=1,2,3,4 and β satisfies
where τ= τ(x)is a scalar function on U.
(2)φ=φ(b2,s)satisfies(3.1)and
Moreover,β satisfies
Moreover,β satisfies
where θ:= θiyiis a 1-form which is perpendicular to β. ξi:= ξi(b2),i=1,2 and
ProofBy(2.2)and(2.5),we can show that a general(α,β)-metric is a Douglas metric if and only if
To simplify the computations,we take the following coordinate transformation:(s,ua)→(yi)by
Then,
Let
Because expression(3.8)involves rij,sijetc,one needs the following expression:
Note that s1=bs11=0.Plugging the above identities into(3.8),we get a system of equations in the following form
where Aijand Bijare polynomials in ya.We must have
Let
For i=1,j=a,by(3.8),we get
From(3.9)and(3.10),then(3.11)is equivalent to
and
For i=a,j=b,by(3.8)we get
It follows from(3.9)and(3.10)that(3.14)is equivalent to
and
Let
Then,(3.16)can be rewritten as
By(3.17),it is easy to show that
We also get
where
Taking s=b in(3.20),we find that the ratio of r11and?B is a function of b2.Hence,suppose thatplugging them into(3.20)yields
Let s=0,then(3.21)is reduced to
where λ3(b2)= χ(b2,0)and λ4(b2)= Ψ(b2,0).Plugging(3.22)into(3.21)yields
By(3.23),there exists a scalar function λ5(b2)such thatIn this case,(3.23)is reformed as
Inserting Ψ and χ into(3.25)yields
By now,we have obtained
Next,we are going to obtain ra1and sa1from(3.13)and(3.15).Let s=0 in(3.15),we obtain
where ε1= ε1(b2).Plugging(3.28)into(3.15)yields
Taking s=b in(3.29),we have
where ε2:= ε2(b2)and ε3:= ε3(b2).Inserting(3.30)into(3.29)yields
By(3.31),we divide the problem into three cases:
(1)ra1=0,sa1=0;
(2)ra1=k(b2)sa1,sa1/=0;
(3)sa1=0,ra1/=0.
Case 1ra1=0,sa1=0.
In this case,by(3.13)and(3.15),it is obtained thatBy sa1=0 and(3.18),then β is closed.From ra1=0 and(3.27),it follows that
By the above equality and β being closed,we obtain(3.2).By(3.26),we obtain(3.1).
Case 2ra1=k(b2)sa1,sa1/=0.
Because sa1/=0,β is not closed.Moreover,we have obtained
Noting that sijis skew symmetric,by(3.32),we obtain(3.5).In this case,ra1=?;Let us summarize what we have proved so far
By(3.33),we have(3.4).
By(3.31)and sa1/=0,we have
where c1= ε2?kε3and c2:=(1?k)ε1.From(3.34),we claim that k/=1.If k=1,then(3.34)is reduced to bQ=c1s.In this case,φ=1+c1bs2.Hence,F is Riemannian.Plugging Q and R into(3.34),we rewrite it as
whereμ1:=c2b2,μ2:=c1?c2.
Differentiating(3.35)with respect to s yields
Contracting(3.13)with yayields
Plugging ra1=ksa1into(3.30)yields
By(3.38),we get
Plugging ra1=ksa1and(3.39)into(3.37)yields
Because the coefficients of the right side in(3.40)is independent of s,if(3.40)holds,then by(3.34),we obtain
where μ3:=and λ3:= λ3(b2).
By making using of(3.35)and(3.36),we have
Plugging(3.42)into(3.41)yields
We claim that k+1+2μ1?μ3b2=0 and 2μ2+μ3=0.If k+1+2μ1?μ3b2/=0,by solving(3.43),we obtain
where g(b2)is an arbitrary smooth function.Hence,the corresponding general(α,β)-metric is of Randers type.This case is excluded in the assumption of Theorem 1.1.If k+1+2μ1?μ3b2=0 and 2μ2+μ3/=0,then we obtain
At this time,the corresponding general(α,β)-metric is of Kropina type.This case is excluded because this class of metrics are singular.Hence,
Case 3sa1=0,ra1/=0.
In this case,for rij,we have obtained
Therefore,by(3.45),we have
where θ := θiyiis a 1-form,which is perpendicular to β.By sa1=0 and(3.18),then β is closed.Hence,it follows from(3.46)that we obtain(3.7).
Because sa1=0 and ra1/=0,by(3.31),we obtain
Plugging R into(3.47)yields
where ξ1:= ε1and.Differentiating(3.48)with respect to s yields
Plugging sa1=0 into(3.13)yields
Contracting(3.50)with respect to yayields
Plugging sa1=0 into(3.30)yields
By(3.52),we conclude that
Plugging(3.53)into(3.51)yields
By(3.47),the above equality is reduced to the following
Because the right side in(3.54)is independent of s,there exists a function η(b2)such that
We rewrite(3.55)as
where ξ3:= η ? ε1.By(3.48)and(3.49),we obtain
Plugging(3.57)into(3.56)yields
We claim that 1?2b2ξ1?b2ξ3=0 and ξ3?2b2ξ2=0.If 1?2b2ξ1?b2ξ3/=0,then by solving(3.58),we obtain
where g(b2)is an arbitrary smooth function.Hence,the corresponding general(α,β)-metric is of Randers type.This case is excluded in the assumption of Theorem 1.1.If 1?2b2ξ1?b2ξ3=0 and ξ3? 2b2ξ2/=0,then we obtain At this time,the corresponding general(α,β)-metric is of Kropina type.This case is excluded because this class of metrics are singular.Therefore,we obtain
By(3.18),we have the following result.
Remark 3.1If a general(α,β)-metric is a Douglas metric,then the Rank of skew symmetric matrix(sij)is not greater than 2.
In this section,we show the sufficient conditions for a general(α,β)-metric to be a Douglas metric.
Lemma 4.1Let F= αφ(b2,s),s=,be a regular general(α,β)-metric on an open subset U ? Rn(n ≥ 3).Suppose that β is not parallel with respect to α and F is not of Randers type.Then,F is a Douglas metric if one of the following cases holds:
(1)φ=φ(b2,s)satisfies
where ki:=ki(b2),i=1,2,3,4 and β satisfies
where τ= τ(x)is a scalar function on U.
(2)φ=φ(b2,s)satisfies(4.1)and
where k:=k(b2)/=1,μi:= μi(b2),i=1,2 and μ1+μ2b2= ?(k+1).Moreover,α and β satisfies
whereμ1+μ2b2=(k+1)and k=k(b2).In this case,dβ/=0.
(3)φ=φ(b2,s)satisfies(4.1)and
and β satisfies
where θ := θiyiis a 1-form which is perpendicular to β. ξi:= ξi(b2),i=1,2 and ξ1+ξ2b2=
ProofAssume that case 1 holds.By(4.2),we have
Plugging(4.8)into(2.5)yields
where χ :=R? 2RΨb2? sΠ and P:= ατ[(k1+k2s2)Θ+(2ΘRb2+s?)(k1+k2b2)].
Inserting(4.1)into(4.9)yields
Thus,it follows from(2.1)and(4.10)that F is a Douglas metric.
Assume that case 2 holds.By(4.4)and(4.5),we obtain
Inserting(4.1),(4.3),(4.11),(4.12)and(3.42)into(2.5)yields
where we have used the first formula of(3.44)and P:= ατ[(k1+k2s2)Θ+(2ΘRb2+s?)(k1+Hence,from(2.1)and(4.13),then F is a Douglas metric.
Assume that case 3 holds.From(4.7),we obtain
Plugging(4.14)and(4.15)into(2.5)yields
where P:= ατ[(k1+k2s2)Θ +(2ΘRb2+s?)(k1+k2b2)]+(2sΘ +b2?)θ.
Plugging(4.1),(4.6),and(3.57)into(4.16)yields
where we have used the first formula of(3.59).Hence,by(2.1)and(4.17),then F is a Douglas metric.
AcknowledgementsThe author would like to thank Professor Changtao Yu for his helpful discussion and the valuable comments.
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Acta Mathematica Scientia(English Series)2018年2期