 Research Article
 Open Access
 Published:
FirstOrder Twistor Lifts
Journal of Inequalities and Applications volume 2010, Article number: 594843 (2010)
Abstract
The use of twistor methods in the study of Jacobi fields has proved quite fruitful, leading to a series of results. L. Lemaire and J. C. Wood proved several properties of Jacobi fields along harmonic maps from the twosphere to the complex projective plane and to the three and fourdimensional spheres, by carefully relating the infinitesimal deformations of the harmonic maps to those of the holomorphic data describing them. In order to advance this programme, we prove a series of relations between infinitesimal properties of the map and those of its twistor lift. Namely, we prove that isotropy and harmonicity to first order of the map correspond to holomorphicity to first order of its lift into the twistor space, relatively to the standard almost complex structures and . This is done by obtaining firstorder analogues of classical twistorial constructions.
1. Introduction
Harmonic maps are mappings between Riemannian manifolds and which extremize the energy functional
Letting denote the tangent bundle of , one can (locally) characterize harmonic maps as solutions of the nonlinear secondorder differential equation
where denotes the tension field of ,
A bibliography can be found in [1] and for some useful summaries on this topic, see [2, 3].
The infinitesimal deformations of a harmonic map are called Jacobi fields. More precisely, let be a smooth map and denote by the set of smooth sections of the pullback bundle . If is harmonic and is a vector field along it, is said to be a Jacobi field (along ) if it satisfies the linear Jacobi equation, where the Jacobi operator is defined by
Here, is the Laplacian on ,
and, letting denote the curvature tensor of ,
Jacobi fields are characterized as lying in the kernel of the second variation of the energy functional. Indeed, if is a twoparameter variation of a harmonic map , then, writing and , the Hessian of is the bilinear operator on given by
so that a Jacobi field (along ) is characterized by the condition
A Jacobi field is called integrable if it is tangent to a deformation through harmonic maps. In [4, 5], the question of whether all Jacobi fields are integrable is answered for the case where the domain is the twosphere and the codomain the twodimensional complex projective space or the three and fourdimensional sphere. This was done by relating the deformations of the map associated with the Jacobi field and those of the twistor lift of the map. More precisely, given an oriented evendimensional manifold , we can construct its (positive) twistor space . This manifold admits two natural almost complex structures and . Given a map from a Riemann surface , harmonicity is intimately related with the existence of a holomorphic lift , whereas isotropy is related with the existence of a holomorphic lift (see [6]). On the other hand, Jacobi vector fields induce families of maps which are harmonic to firstorder and, in some cases, isotropic to first order. The translation of these first order properties in terms of twistor lifts plays an important role on the study of the Jacobi fields and we shall exhibit how this translation can be established in general.
This work is divided as follows: in the next two sections, we recall some wellknown results concerning twistor lifts of harmonic and isotropic maps. In Section 4, we show how this constructions generalize to their parametric versions and examine more closely the construction when the codomain is a fourdimensional manifold. We leave to the last section some technical proofs.
2. The Setup
2.1. Twistor Spaces
Let be an oriented evendimensional Euclidean space, equipped with a metric . A Hermitian structure on is with and for all . We say that is positive if there is a positive basis of of the form and negative otherwise. The set of all positive Hermitian structures (resp., all negative Hermitian structures) on is denoted by (resp., ). The Lie group acts transitively on by the formula and the isotropy subgroup at is given by
Letting and , we easily conclude that
In particular, the tangent space of at is given by . In this vector space, we can introduce a complex structure defining
When equipped with , the manifold is a complex manifold.
Given with a Hermitian structure , we can consider on its and parts given as usual by
These are isotropic subspaces, in the sense that for all in (or in ). Associating an Hermitian structure on with its space gives a 1–1 correspondence between Hermitian structures and maximal isotropic subspaces. We say that a maximal isotropic subspace is positive if the corresponding orthogonal complex structure is positive and we denote the set of all such subspaces by .
Let be an oriented evendimensional vector bundle over a manifold equipped with a connection and a parallel metric . Then, we may take the bundle
whose fibre at is precisely . If is Riemann surface, the vector bundle has vanishing part of its curvature tensor and therefore admits a unique structure as a holomorphic bundle over , by a wellknown theorem of Koszul and Malgrange [7]. This induces a holomorphic structure on the bundle for which a section of is holomorphic if and only if [6] (see [8]).
Let be an oriented Riemannian manifold with dimension . We call (positive) twistor space of the bundle whose fibre at is precisely ; that is,
The Riemannian connection on induces a splitting of the tangent space to into vertical and horizontal parts, . Namely, if denotes the canonical projection defined by , then
With respect to this decomposition, maps isomorphically into and allows to define an almost complex structure on as the pullback of on . Together with (2.3), this allows to define two almost complex structures and on by the formulae
When equipped with , is never a complex manifold; as for , it is integrable if and only if is conformally flat () or antiselfdual () (for more details, see [6, 9, 10]; a discussion on this topic can also be found in [11] and references therein).
Notice that is a holomorphic map for any of these complex structures; that is, for or , one has that
Definition 2.1.
Let and be two almost complex manifolds. Let be a decomposition of into stable subbundles; that is, and . We shall call such a decomposition a stable decomposition. Let be a smooth map. We shall say that is holomorphic if
Analogously we define holomorphic maps.
A smooth map is holomorphic if and only if it is both and holomorphic for some, and so any, stable decomposition . Taking , the decomposition is clearly stable for both the almost complex structures and on .
Remark 2.2.
We can easily introduce a metric on the twistor space : let and consider the tangent space at this point, . We know that we have the identifications and . To get a metric on , transport the metric from that on ; that is, , for all ,, where denotes the metric on at . For the vertical space , we can consider the restriction of the metric on the space . Finally, we declare and to be orthogonal under the metric ; that is, , for all , . With this metric, the decomposition is orthogonal and stable (), () are almost Hermitian manifolds and the projection map is a Riemannian submersion.
2.2. Conformal and Isotropic Maps
Given a smooth map , is said to be weakly conformal at if there is with
If , then is said to be a regular point (of ) and the map is called conformal at . Moreover, a map which is conformal (resp., weakly conformal) at all points is said to be a conformal map (resp., a weakly conformal map).
If is an almost Hermitian manifold, any holomorphic (resp., antiholomorphic) map is (weakly) conformal as it maps to (resp., to ). A stronger property than conformality is isotropy: if is a smooth map from a Riemann surface, is isotropic if [12]
where . Actually, the condition (2.13) can be weakened to
To check this, establish an induction on : if the result is trivial. Assuming now that (2.14) implies (2.13) for all and taking with , we may assume without loss of generality that , and we get
Since and , both terms in the above expression vanish.
Moreover, letting in (2.13), it is easy to check that an isotropic map from a Riemann surface is a (weakly) conformal map.
Let be a smooth map from a Riemann surface . We shall say that is an umbilic point (of ) if is a linearly dependent set. If is such that all points are umbilic, we shall say that is totally umbilic (see [6]).
3. Nonparametric Twistorial Constructions
The following are wellknown twistorial constructions [13] (see also [6]).
Theorem 3.1.
If is holomorphic, the projection map is isotropic. Conversely, if is a conformal totally umbilic immersion, there is (locally) a holomorphic map such that .
If is holomorphic, the projection map is harmonic. Conversely, if is a conformal harmonic map, there is (locally) a holomorphic map such that .
We shall sketch the proof of this result. We start by noticing that given a smooth map obtained as the projection of , , without requiring further conditions à priori on , nothing guarantees that is holomorphic relatively to the induced almost Hermitian structure on ; if it is, we shall say that the structure is strictly compatible with (or that the map is a strictly compatible twistor lift of ). Such a structure can exist if and only if is isotropic: in other words, if and only if is (weakly) conformal. If preserves but does not necessarily render holomorphic, we shall say that (or the map ) is compatible with .
If is given as the projection of an holomorphic map ( or ), then is holomorphic with respect to the induced almost Hermitian structure on :
In particular, is (weakly) conformal. Moreover, the above equivalence shows that any strictly compatible lift of is holomorphic. On the opposite direction, let be a conformal map. Let denote the normal bundle of in . We may decompose the connection on into its tangent and normal parts, . Hence, on we have a metric and connection inherited from those on . Moreover, we may take the bundle over which has the holomorphic structure. Since is conformal, we may transfer the Hermitian structure of into . Hence, we have a natural map
Taking any holomorphic section of and composing with , we obtain a strictly compatible twistor lift of . Since any such lift is holomorphic, we have just proved the following result.
Proposition 3.2.
Given , is conformal if and only if is (locally) the projection of an holomorphic map .
To proceed, we need the following result [13].
Proposition 3.3.
Let and let . Take the section of corresponding to . Then, the map is holomorphic if and only if is holomorphic with respect to and
The map is holomorphic if and only if is holomorphic with respect to and
Let be a holomorphic map and let be the corresponding section of . If is the projection map, we have that is holomorphic. In particular, . Since is holomorphic, we can write
and, inductively, it follows that for all so that is isotropic. Conversely, let be a conformal totally umbilic map. In this case, we consider the manifold equipped with the opposite holomorphic structure and again construct the holomorphic bundle . For this structure, a section is holomorphic if and only if . Since is conformal, we may define the map as in (3.2). On the other hand, because is totally umbilic, we know that lies in the span of ; in other words, if is any almost Hermitian structure on strictly compatible with , lies in . We may therefore conclude that is holomorphic. As a matter of fact, given a holomorphic section of and considering , we have that . We may write
since and lie in . In particular, and so
and therefore we conclude that is holomorphic.
With a similar argument, let be a holomorphic map and let be the section of corresponding to . Then, taking the projection map , we get
Since is holomorphic, lies in . From (iv) in Proposition 3.3, we conclude that lies in and therefore has vanishing part. From the reality of , we have that and so is harmonic. Conversely, if is conformal and harmonic, we may take the bundle and the map in (3.2). Since is harmonic, is holomorphic with respect to the and structures [6, 8]: letting denote a holomorphic section of and the composition , then, and
Since is harmonic, . On the other hand, since is holomorphic, . Finally,
shows that and so that is holomorphic. Therefore, taking any holomorphic section of and composing with give a holomorphic lift of .
Remark 3.4.
Notice that to guarantee the existence of the holomorphic lift for , the important fact was that belongs to the part of for any almost Hermitian structure strictly compatible with . This is guaranteed if is a totally umbilic map, but it is not strictly necessary. For instance, if is an isotropic map, the vectors , span an isotropic subspace. If this vectors are linearly independent, taking this space as the space of , then is a holomorphic lift of , although may be a map into ; on the other hand, if is totally umbilic, then we may take either as the unique strictly compatible map into or into and both these maps are holomorphic.
4. FirstOrder Twistorial Constructions
4.1. Harmonicity and Isotropy to First Order
Let denote an interval of the real line containing . Given a (family of) map(s) , , we say that is harmonic to first order if
where and .
Let be a harmonic map, a vector field along , and a oneparameter variation of . We say that is tangent to if . The following result is a key ingredient in what follows [4]:
Proposition 4.1.
Let be a harmonic map between compact manifolds and . Let be a vector field along and a oneparameter variation of tangent to . Then,
In particular, is Jacobi if and only if any tangent oneparameter variation is harmonic to first order.
We have seen in Theorem 3.1 that harmonicity was not enough to establish a relation with possible twistor lifts of a map conformality and was also a key ingredient, as maps obtained as projections of twistorial maps must be holomorphic with respect to some almost Hermitian structure along the map. On the other hand, when the domain is the sphere, harmonicity implies (weak) conformality or even isotropy, the last case occurring if the target manifold is itself also a sphere or the complex projective space [12, 14].
Let be a Riemann surface and a smooth map. The map is said to be conformal to first order if [15]
Analogously, is said to be isotropic to first order,
As for the nonparametric case, one can prove [16] that condition (4.4) can be weakened to the following
As in the nonparametric case, harmonicity to first order implies conformality to first order for maps defined on the twosphere and even isotropy when the codomain is itself a real or complex space form [15].
4.2. Twistorial Constructions
As we have seen, Jacobi fields induce variations that are harmonic (and, in some cases, conformal or even isotropic) to first order. On the other hand, in Section 2 we have seen that conformality, harmonicity, and isotropy of the map correspond to , and holomorphicity of the twistor lift . As we shall see, these results do have a firstorder version as follows. We start with a definition.
Definition 4.2.
Let be an almost complex manifold and an almost Hermitian manifold. Given a smooth map , we say that is holomorphic to first order if is holomorphic and
where is the LeviCivita connection on induced by the metric . Moreover, if is a stable decomposition of , orthogonal with respect to , we shall say that is holomorphic to first order if is holomorphic and
where is the restriction of to . Changing to gives the definition of holomorphicity to first order.
In contrast with the nonparametric case, it is not obvious that holomorphicity to first order implies holomorphicity to first order. As a matter of fact, from (4.6), it only follows that . However, we do have the following.
Lemma 4.3.
Let be a smooth map and let be a stable decomposition of , orthogonal with respect to . Then, is holomorphic to first order if and only if is both and holomorphic to first order.
Proof.
Assume that is holomorphic to first order. Then, is holomorphic. As for (4.7), letting denote an arbitrary section of and its projection into , we have
which is true since is holomorphic to first order. Hence, is holomorphic to first order. Changing to shows that is holomorphic to first order. The converse also follows using analogous arguments.
Remark 4.4.
The importance of choosing the LeviCivita connection on is illusory. In particular, we can define the concept of holomorphicity to first order (or,holomorphicity to first order) for maps defined between almost complex manifolds, not necessarily equipped with any metric.
Indeed, let be a holomorphic to firstorder map with respect to , so that (4.6) holds. Let denote a (local) frame for . Then,
Since is holomorphic to first order, is holomorphic, the above equation is equivalent to
Now, since holomorphicity does not depend on the chosen connection, we can deduce that holomorphicity with respect to reduces to the same condition (4.10). Thus, being holomorphic to first order does not depend on the chosen connection. For (resp., ) holomorphicity to first order, we use similar arguments, replacing for a horizontal (resp., vertical) frame.
4.3. The Holomorphic Case
In the nonparametric case, given a conformal map , we can always find a lift such that is holomorphic with respect to . In other words, (locally defined) strictly compatible lifts always exist. In general, this lift may not be or holomorphic but it is holomorphic. Let be a variation of , conformal to first order. Then, if a lift to the twistor space that makes holomorphic for all small exists, is necessarily conformal for all small , which may not be the case. So, even if conformality is preserved to first order, there might be no strictly compatible twistor lift for all ; hence, we should relax the condition on conformality. We shall say that a twistor lift of a conformal to first order map is compatible to first order (with ) if
We start with a technical lemma, whose proof the reader can find in Section 5.
Lemma 4.5.
Let be a conformal to firstorder map. Let be a twistor lift compatible to first order with . Then for all there is a function and a vector field with , and , such that
In particular, is holomorphic to first order in the sense that
Lemma 4.6.
Let be holomorphic to first order. Then, is holomorphic to first order.
Proof.
Since is holomorphic to first order, . Therefore, for all , since is holomorphic,
Since is a holomorphic Riemannian submersion, the above equation can be written as
Hence, for all , using the fact that is holomorphic,
showing that and concluding our proof.
Proposition 4.7.
Let be holomorphic to first order map. Then, the projected map is conformal to first order. Conversely, let be a conformal to first order map. Then there is a (local) holomorphic to first order map which is compatible to first order with .
Proof.
Take an holomorphic to first order map and let . We know that is conformal (Proposition 3.2). As for the firstorder variation, using the preceding Lemma 4.6,
Using similar arguments, we can show that
concluding the first part of the proof.
Conversely, let be any twistor lift of compatible to first order. Then, there is a function and a vector field verifying the conditions on Lemma 4.5 with
Now, is holomorphic to first order if and only if is holomorphic (which follows from Proposition 3.2) and (4.6) holds. Using the same argument as in Lemma 4.3, (4.6) is equivalent to
But
from the given conditions on and and thus concluding the proof.
4.4. The Holomorphic Case
Next, we give a useful characterization for maps to be holomorphic to first order ( or ), whose proof is in Section 5 (compare with Proposition 3.3).
Lemma 4.8.
Let be a smooth map and let be its projection. Then, is holomorphic to first order ( or ) if and only if
or
From the preceding lemma, we can also deduce the following.
Lemma 4.9.
Let be a map holomorphic to first order and consider the projected map . Then for all there is with
Proof.
For , we have so that using (4.22) we have
Taking , we obtain the result. To establish an induction, assume now that the result is valid for ; that is, there is such that
Taking and noticing that ,
as is antisymmetric on the first two arguments. Now, since is holomorphic, (4.23) holds so that there is such that
But the second condition gives , whereas the first holds precisely that , as we wanted to show.
Proposition 4.10 (projections of maps holomorphic to first order).
Let be a map holomorphic to first order, where is any Riemann surface. Then, the projection map is isotropic to first order.
Notice that we could replace with , as real isotropy (to first order) does not depend on the fixed orientation on .
Proof.
That is isotropic follows from the nonparametric case. Therefore, we are left with proving that
Using Lemma 4.9, for fixed , choose with and . Then, the conclusion follows from
We now turn our attention to the existence of lifts holomorphic to first order for a given isotropic to firstorder map . Recall that in the nonparametric case such lift exists (see Remark 3.4).
Theorem 4.11.
Let be an isotropic to first order with and linearly independent. Then, there is either a (local) map or a map which is holomorphic to first order and compatible to first order with .
Before proving Theorem 4.11, we give the following lemma, which we prove in Section 5
Lemma 4.12.
Let be as in the preceding Theorem 4.11.
(i)Suppose that the holomorphic lift of is (resp., ). Take the unique positive (resp., negative) almost Hermitian structure on compatible with . Then, taking
we have that
(ii)There are , such that
We are now ready to prove Theorem 4.11.
Proof of Theorem 4.11.
As before, take or the holomorphic lift of . Assume, without loss of generality, that it is . Then, at each take the unique positive almost Hermitian structure compatible with and let us prove that this map is holomorphic to first order. Using Lemma 4.5, is holomorphic to first order and we are left with proving that (4.23) holds. It is enough to prove that there is a basis of for which (4.23) holds. Now, take and where is as in (4.32). Then,
Analogously,
where we have used , , , and . Hence, and satisfy equation (4.23), concluding our proof.
4.5. The Holomorphic Case
We prove the following.
Theorem 4.13.
Let be a map holomorphic to first order. Then, is harmonic to first order (and conformal to first order from Lemma 4.3 and Proposition 4.7).
We first give the following characterization of holomorphic to first order maps.
Lemma 4.14.
Let ( or ) be a smooth map. Then, is a holomorphic to first order map if and only if is holomorphic and
Proof.
If is any smooth map, then [17] . Hence, from (2.9),
Thus, we can rephrase equation (4.37) as
which is the condition for holomorphicity.
Proof of Theorem 4.13.
That is harmonic follows from Theorem 3.1. Hence, we are left with proving that . Since is holomorphic to first order, we deduce that is both and holomorphic to first order (Lemma 4.3). From holomorphicity (4.37), we have
Using Lemma 4.6 and (4.13) together with symmetry of the second fundamental form of , the righthand side of the above identity becomes
so that , concluding the proof.
Theorem 4.15.
Let be a harmonic and conformal to first order map. Then, there is (locally) a map which is holomorphic to first order and with .
Since harmonicity (to first order) does not depend on the orientation on , we could have replaced by in both Theorems 4.13 and 4.15
Proof.
For each consider , bundle over . Since is a Riemann surface, and we can conclude that for each there is a KoszulMalgrange holomorphic structure on . Hence ([16, Theorem .]), there is a smooth section with a KoszulMalgrange holomorphic section of : . So,
equivalently,
Take where is the part on determined by = rotation by + on (). (Notice that as is not conformal we might not get a Hermitian structure by setting ; on the other hand, positive rotation by comes from the natural orientation on imported from via.) Then defines a compatible twistor lift of . Let us check that is holomorphic to first order. That is holomorphic is immediate. From the proof of Proposition 4.7, we deduce that is holomorphic to first order as it is compatible to first order with and the latter is conformal to first order. Hence, we are left with proving that (4.37)
holds. We shall establish this equation by showing that both sides agree when applied to any vector . For that, we consider, in turn, the three cases , , and . The first two have similar arguments so that we prove only the first.
(i). From holomorphicity, we have . On the other hand, as is holomorphic to first order, (4.13) is satisfied. Finally, for all ,
Since is harmonic to first order, our condition follows from
(ii). We now have
Now, the first condition follows from (4.43) since is KoszulMalgrange holomorphic for each . As for the second, letting denote its lefthand side, we shall prove that for all . We do this by establishing the three cases , and (since the first two cases have similar arguments, we prove only the first).
( )When we have
( )Let . Then,
The first term on the right side of the above equation vanishes as lies in , whereas the second is zero from holomorphicity, concluding our proof.
4.6. The 4Dimensional Case
Theorem 4.16.
Let be harmonic and isotropic to firstorder map and with and being linearly independent. Then, (locally) there is either a map or a map which is simultaneously and holomorphic to first order and with . Conversely, if (or ) is and holomorphic to first order, the projected map is harmonic and isotropic to first order.
Proof.
The converse is obvious from Proposition 4.10 and Theorem 4.13. As for the first part, in Theorem 4.11 we saw that we can lift the map to a map holomorphic to first order. Moreover, this lift could be defined as the unique positive or negative almost complex structure compatible with . On the other hand, in Theorem 4.15 we have seen that there is a map holomorphic to first order with and for which is compatible. From the comment after Theorem 4.15, there is also a twistor lift of into . Therefore, from the dimension of , we conclude that the lifts constructed in both cited results are the same and, therefore, simultaneously and holomorphic to first order.
We would now like to guarantee the uniqueness to first order of our twistor lift. Before stating such a result, we start with a lemma, proved in Section 5.
Lemma 4.17.
Let be a map holomorphic to first order. Consider the twistor projection and the vectors and defined by
so that
Then, for all for which and are linearly independent
Notice that in Lemma 4.12, we were given and defined the twistor lift as the unique lift compatible with . Now, we are given the twistor map but nothing guarantees that projecting the map to makes compatible; that is, may not preserve .
Proposition 4.18.
Let be two holomorphic to firstorder maps such that and the variational vector fields induced on are the same; that is, writing , , we have . Then, at all points for which and are linearly independent, writing , , we have .
Proof.
Let () denote the projection maps. From our hypothesis, it follows that . Hence, the only thing left is to prove that the vertical parts coincide. Now, from the proof of Lemma 4.14, so that our result follows if . We prove this identity showing that for all , where
We consider the four possible cases for ; namely, when is equal to , , or , where and are as in the preceding lemma (notice that, since , then and ).
(i)When ( uses similar arguments), we have
where we have used Lemma 4.17, as well as the fact that and .
(ii)Taking ( uses similar arguments), we have
But
As , we deduce ; analogously, so that .
Hence, the twistor lifts constructed in Theorem 4.16 are unique to first order, in the sense that the vector field induced on (or ) by the map , depends only on the initial projected map and on the Jacobi field along . Moreover, taking the sphere or the complex projective plane, letting be a harmonic map, and a Jacobi field, isotropy to first order is immediately guaranteed. Hence, the previous construction allows a (local) unified proof of the twistor correspondence between Jacobi fields and twistor vector fields that are tangent to variations on which are simultaneously and holomorphic (infinitesimal horizontal holomorphic deformations in [5]). We can also conclude which different properties (namely, conformality, real isotropy or harmonicity) are related with those of the twistor lift (resp., , or holomorphicity).
5. Additional Proofs
Proof of Lemma 4.5.
Since is compatible to first order, preserves for all . Hence, there are and such that
Take . Since, at , we deduce and . Now, since and form an orthogonal basis for , we have
Differentiating with respect to at the point , the above identity yields
Computing the inner product of (5.1) with and using the fact that vanishes for all , we get . Hence,
and we deduce , as and (5.3) hold. Using (5.1) again, we can now write
which vanishes since for all , the second and last terms cancelling as is holomorphic and is conformal to first order. Analogously, vanishes so that . For the orthogonal part, taking ,
showing that and concluding the proof.
Proof of Lemma 4.8.
We shall do the proof only for the case, with the case being similar. Assume that is holomorphic to first order. Then, using Lemmas 4.3 and 4.6, is holomorphic to first order. On the other hand, satisfies equation (4.37), which implies that for all
Take in . Let be such that and write . Then,
Moreover, since is holomorphic, and
finishing the first part of our proof.
Conversely, suppose that (4.22) and (4.23) hold. Take and with and . Then
and we can now easily conclude that (5.7) holds. Together with the fact that is holomorphic (Proposition 3.3), we can conclude that (4.37) is verified. As for the horizontal part, we have that
as is holomorphic. Since (4.22) holds, the last condition is trivially satisfied and we can conclude that our map is holomorphic to first order, as desired.
Proof of Lemma 4.12.

(i)
Since is isotropic to first order, , equivalently, . Thus, we have
(5.12)
Similarly, is equivalent to and implies
As is compatible with , Lemma 4.5 guarantees that is holomorphic to first order. On the other hand, since and are linearly independent, we deduce that , and form a basis for . Hence, (4.33) will be satisfied if and only on evaluating the inner product of and with which one of these four vectors one obtains the same result. We shall only prove for the first and fourth vectors, the other two cases being similar.
(a)Since ,
(b)Using (5.13),

(ii)
We know that , , , span . Hence, there are , and with
(5.16)
where since . Therefore,
Now, using the fact that is isotropic to first order, we have
Together with
we deduce
Similarly, from
we have
Equations (5.20) and (5.22) imply that either (which is absurd as does not lie in ) or and consequently (4.34) holds, as wanted.
Proof of Lemma 4.17.
That follows from the proof of Proposition 4.7. Since , and are linearly independent vectors, we can deduce that , , and form a basis for for is a neighbourhood of . On the other hand, as is the projection of a map holomorphic to first order, we know that it must be isotropic to first order from Proposition 4.10. Hence,
The argument to establish the second and third identities in (4.52), will now be similar to the one in Lemma 4.12(i).
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Acknowledgments
The author is grateful to Professor J. C. Wood for helpful comments and stimulating discussions during the preparation of this work. The author would also like to thank the referee for valuable comments.
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Simões, B.A. FirstOrder Twistor Lifts. J Inequal Appl 2010, 594843 (2010). https://doi.org/10.1155/2010/594843
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DOI: https://doi.org/10.1155/2010/594843
Keywords
 Riemann Surface
 Complex Manifold
 Twistor Space
 Holomorphic Section
 Hermitian Structure