complex if ication

7/31/2019 Complex if Ication

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COMPLEXIFICATION

KEITH CONRAD

1. Introduction

We want to describe a procedure for enlarging real vector spaces to complex vector spacesin a natural way. For instance, the natural complex analogues of R n , Mn (R ), and R [X ]are C n , Mn (C ) and C [X ].

Why do we want to complexify real vector spaces? One reason is related to solvingequations. If we want to prove theorems about real solutions to a system of real linear

equations or a system of real linear differential equations, it can be convenient as a rststep to examine the complex solution space. Then we would try to use our knowledgeof the complex solution space (for instance, its dimension) to get information about thereal solution space. In the other direction, we may want to know if a subspace of C nwhich is given to us as the solution set to a system of complex linear equations is also thesolution set to a system of real linear equations. We will nd a nice way to describe suchsubspaces once we understand the different ways that a complex vector space can occur asthe complexication of a real subspace.

We will give two descriptions of the complexication process, rst in terms of a two-fold direct sum (Section 2) and then in terms of tensor products (Section 3). The tensorproduct viewpoint is the more far-reaching one, but seeing how the direct sum method of complexication expresses everything may help convince the reader that there is nothing

unexpected about this use tensor products. Moreover, idiosyncratic aspects of the directsum construction will turn out to be completely natural features of the tensor productconstruction. After comparing the two constructions, we will see (Section 4) how to usea special structure on a complex vector space, called a conjugation, to describe the realsubspaces which have the given complex vector space as its complexication.

References on this topic are [1, pp. 7981], [2, 77], and [3, pp. 262263]. In [1] and [3]tensor products are used, while [2] uses direct sums.

2. Complexifying with Direct Sums

Let W be a real vector space. To create from W a naturally associated complex vectorspace, we need to give a meaning to ( a + bi)w, where a + bi C and w W . Whatever itmight mean, wed like to have

(a + bi)w = aw + biw = aw + ibw,

and since there is no given meaning to ibw (that is, multiplying elements of W by i has nodenition since W is a real vector space), aw + ibw should be thought of as a formal sumof aw and bw with the i factor keeping them apart. A legitimate way to make sense of thisis to treat aw + ibw as the ordered pair ( aw,bw), and that leads to the following denition.

1


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2 KEITH CONRAD

Denition 2.1. The complexication of a real vector space W is dened to be W C =W W , with multiplication law ( a + bi)(w1, w2) = ( aw1 bw2, bw1 + aw2), where a and bare real.

This rule of multiplication is reasonable if you think about a pair ( w1, w2) in W W asa formal sum w1 + iw2:

(a + bi)(w1 + iw2) = aw1 + aiw 2 + biw1 bw2 = ( aw1 bw2) + i(bw1 + aw2).

In particular,

(2.1) i(w1, w2) = ( w2, w1).

It is left to the reader to check that W C is a complex vector space by the multiplicationrule in Denition 2.1 ( e.g. , z(z (w1, w2)) = zz (w1, w2)). Since

i(w, 0) = (0 , w),

we have

(2.2) (w1, w2) = ( w1, 0) + (0 , w2) = ( w1, 0) + i(w2, 0).

Thus elements of W C formally look like w1 + iw2, except iw2 has no meaning while i(w2, 0)does: it is (0, w2).

The real subspaces W {0} and {0}W of W C both behave like W , since addition iscomponentwise and a(w, 0) = ( aw, 0) and a(0, w) = (0 , aw ) when a is real. The R -linearfunction w (w, 0) will be called the standard embedding of W into W C . So we treatW {0} as the official copy of W inside W C and with this identication made we canregard W C as W + iW using (2.2).

Example 2.2. Taking W = R , its complexication R C is the set of ordered pairs of real numbers ( x, y ) with ( a + bi)(x, y) = ( ax by,bx + ay). Since (a + bi)(x + yi) =

(ax by) + ( bx + ay)i, R C is isomorphic to C as C -vector spaces by ( x, y) x + yi.Example 2.3. The complexications of R n , Mn (R ), and R [X ] are isomorphic to C n ,Mn (C ), and C [X ], by sending an ordered pair ( w1, w2) in R n R n , Mn (R ) Mn (R ), orR [X ]R [X ] to w1 + iw2 in C n , Mn (C ), or C [X ].

For instance, we can identify the complexication ( R 2)C with C 2 (as complex vectorspaces) by ( w1, w2) w1 + iw2 R 2 + iR 2 = C 2, and this sends the basis vectors 10 and

01 of R

2 (R 2)C to 10 and

01 in C

2, which are the standard basis vectors of C 2 as acomplex vector space. More generally, the identications of ( R n )C , Mn (R )C , and R [X ]Cwith C n , Mn (C ), and C [X ] turn every real basis of R n , Mn (R ), and R [X ] (viewed insidetheir complexications by the standard embedding) into a complex basis of C n , Mn (C ),and C [X ].

Theorem 2.4. If W = 0 then W C = 0 . If W = 0 and {e j } is an R -basis of W then {(e j , 0)} is a C -basis of W C . In particular, dim C (W C ) = dim R (W ) for all W .

Proof. That W being zero implies W C is zero is easy. Now take W = 0 with basis {e j }.For ( w1, w2) W C , writing w1 and w2 as R -linear combinations of the e j s shows every

element of W C is an R -linear combination of the ( e j , 0)s and (0 , e j )s. Since (0, e j ) =i(e j , 0), using C -linear combinations we can write every element of W C in terms of thevectors ( e j , 0). Therefore {(e j , 0)} is a C -linear spanning set of W C . To show it is linearly


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4 KEITH CONRAD

Theorem 2.7. Let W and W be nonzero nite-dimensional real vector spaces. For any R -linear transformation : W W and R -bases {ei } and {e j } of W and W , the matrix for with respect to these bases is the matrix for C : W C W C with respect to the C -bases {(ei , 0)} and {(e j , 0)} of the complexications.

Proof. The columns of the matrix for with respect to the bases {ei } and {e j } are thecoefficients which express the values of on each ek as an R -linear combination of the e j s.Similarly, the columns of the matrix for C with respect to the bases {(ei , 0)} and {(e j , 0)}are the coefficients which express the values of C on each (ei , 0) as a C -linear combinationof the (e j , 0)s. Since C (ei , 0) = ( (ei ), 0), the linear combination of (ei ) in terms of the e j s will give the same linear combination of C (ei , 0) in terms of the ( e j , 0)s. So thematrices of and C with respect to these bases are equal.

Example 2.8. Let A M2(R ) act on R 2. The denition of AC on (R 2)C isAC (w1, w2) = ( Aw1, Aw 2),

and the isomorphism f : (R 2)C C 2 by f (w1, w2) = w1 + iw2 identies AC with thefunction

w1 + iw2 Aw1 + iAw 2on C 2. Since A as a matrix acting on complex vectors satises Aw1 + iAw 2 = A(w1 + iw2),the diagram

(R 2)CA C / /

f

(R 2)C

f

C 2A / /C 2

commutes. This is the sense in which the complexication AC is just the matrix A actingon C 2.

It is straightforward to check from the denitions that if : W W and : W W are R -linear transformations then ( )C = C C , so complexication of linear mapscommutes with composition, and easily (id W )C = id (W C ) .

If U is a real subspace of W , then U C is a complex subspace of W C (check). In particular,associated to any R -linear map : W W are the real subspaces ker W and im W . We also have the complexied C -linear map C : W C W C and its kernel and image,which are C -subspaces of W C and W C . The construction of kernels and images behavesnicely with respect to complexication:

Theorem 2.9. If : W W is R -linear, its complexication C : W C W C has kernel and image

ker(C ) = (ker )C , im(C ) = (im )C .

Proof. Since C (w1, w2) = ( (w1),(w2)), the condition ( w1, w2) ker C is equivalent tow1 and w2 lying in ker , so

ker(C ) = (ker ) (ker ) W W = W C .Thus ker( C ) = (ker )C as a subset of W C .

The image of C is (W ) (W ), which is the complexication of (W ) = im .

Corollary 2.10. If : W W is R -linear, then C is injective if and only if is injective, and C is surjective if and only if is surjective.


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COMPLEXIFICATION 5

Proof. Injectivity of C is equivalent to ker( C ) = (ker )C being 0, and a real vector spacehas complexication 0 if and only if it is 0. If is surjective then im( C ) = (im )C = W C ,so C is surjective. If is not surjective then its image (W ) is a proper subspace of W , so im(C ) = (W ) (W ) is a proper subspace of W W and therefore C is not

surjective.As with most important constructions, we can describe the complexication W C of a

real vector space W by a universal mapping property. We have a standard embeddingW W C , which is an R -linear transformation. Consider now all R -linear transformationsW f V of the particular real vector space W into complex vector spaces V . The standardembedding W W C is just one example, but it is really the most basic one, as the nexttheorem shows.

Theorem 2.11. For any R -linear map W f V from W into a complex vector space V ,

there is a unique C -linear map W Cef

V making the diagram

W C

ef

W

7 7 o o o o o o o o o o o o o

f ' ' O O

O O O O O

O O O O

O O O

V

commute, where the map W W C is the standard embedding.

Proof. This is quite similar to the proof of Theorem 2.6, so we just sketch the idea. Assumingf exists, show it satises

f (w1, w2) = ( f (w1), f (w2))by C -linearity. Now dene f by this formula and check it is C -linear and makes the diagramcommute.

To put this in the language of universal mapping properties, if we take as objects allR -linear maps W f V from the xed R -vector space W to varying C -vector spaces V ,call a morphism from W f

1

V 1 to W f 2 V 2 any C -linear (not just R -linear!) map

V 1f V 2 such that the diagram

V 1

f

W

f 1 8 8 p p p p p p p p p p

p p p

f 2 & & N N

N N N N

N N N N

N N N

V 2

commutes. Then Theorem 2.11 says W W C is an initial object (it admits a uniquemorphism to all other objects), so it is determined up to a unique isomorphism by themapping property in Theorem 2.11.


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6 KEITH CONRAD

3. Complexifying with Tensor Products

While the denition of W C does not depend on a choice of basis for W , it implicitlydepends on a choice of the real basis {1, i} of C instead of other bases. This can be seenin the way multiplication by C scales pairs ( w

1, w

2) W C , which we want to think of as a

formal version of w1 + iw2. Although the universal mapping property in Theorem 2.11 givesthe complexication construction a basis-free meaning (within the framework of mappingreal vector spaces to complex vector spaces), it is also possible to give a completely basis-freeconstruction of the complexication using tensor products. The idea is that W C behaveslike C R W and the complexication C : W C W C of an R -linear map : W W behaves like the C -linear map 1 : C R W C R W . Here are some similaritiesbetween W C and C R W :

(1) There are standard embeddings W W C by w (w, 0) and W C R W byw 1w, and with these embeddings we have W C = W + iW and C R W =W + iW .

(2) For a nonzero real vector space W , any R -basis {e j } of W gives us a C -basis {1e j }

of C R W , so the C -dimension of C R W equals the R -dimension of W . Thislooks like Theorem 2.4.(3) The identications of ( R n )C , Mn (R )C , and R [X ]C with C n , Mn (C ), and C [X ] in

Example 2.3 remind us of the effect of base extension to C of R n , Mn (R ), andR [X ].

(4) In the nite-dimensional case, the matrices for and 1 are equal when usingcompatible bases, just as in Theorem 2.7.

(5) Theorem 2.9 is similar to the formulas

ker(1 ) = C R ker , im(1 ) = C R (im )

for kernels and images under base extension.The constructions of both W C and C R W from W should be two different ways of

thinking about the same thing. Lets make this official.Theorem 3.1. For every real vector space W , there is a unique isomorphism f W : W C C R W of C -vector spaces which makes the diagram

W

} } z z z z

z z z z

$ $ I I I

I I I I

I I

W Cf W / /C R W

commute, where the two arrows out of W are its standard embeddings. Explicitly,

(3.1) f W (w1, w2) = 1 w1 + i w2.

Moreover, if : W W is any R -linear map of real vector spaces, the diagram of C -linear maps

(3.2) W CC

/ /

f W

W C

f W

C R W 1

/ /C R W

commutes.


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COMPLEXIFICATION 7

Proof. Assuming f W exists, we must have

f W (w1, w2) = f W ((w1, 0) + i(w2, 0))= f W (w1, 0) + if W (w2, 0)

= 1 w1 + i(1 w2)= 1 w1 + i w2.

Now if we dene f W by this last expression, it is easy to see f W is R -linear. To see f W isin fact C -linear, we compute

f W (i(w1, w2)) = f W ( w2, w1) = 1 ( w2) + i w1 = 1w2 + i w1and

if W (w1, w2) = i(1 w1 + i w2) = i w1 + ( 1) w2 = 1w2 + i w1.

To show f W is an isomorphism, we write down the inverse map C R W W C . Wed like1w to correspond to ( w, 0), so we expect zw = z(1w) should go to z(w, 0). With this

in mind, we construct such a map by rst letting C W W C by ( z, w) z(w, 0). This isR -bilinear, so it induces an R -linear map gW : C R W W C where gW (z w) = z(w, 0)on elementary tensors. To show gW is C -linear, it suffices to check gW (zt ) = zgW (t) whent is an elementary tensor, say t = z w:

gW (z(z w)) = gW (zz w) = zz (w, 0), zgW (z w) = z(z (w, 0)) = zz (w, 0).

Finally, to show f W and gW are inverses, rst we have

gW (f W (w1, w2)) = gW (1 w1 + i w2) = ( w1, 0) + i(w2, 0) = ( w1, 0) + (0 , w2) = ( w1, w2).

To show f W (gW (t)) = t for all t C R W , it suffices to verify this on elementary tensors,and this is left to the reader. The commutativity of (3.2) is also left to the reader.

Since W C and C R W are both the direct sum of subspaces W and iW (using thestandard embedding of W into W C and C R W ), this suggests saying a complex vectorspace V is the complexication of a real subspace W when V = W + iW and W iW = {0}.For example, in this sense C n is the complexication of R n . The distinction from thepreceding meaning of complexication is that now we are talking about how a pre-existingcomplex vector space can be a complexication of a real subspace of it, rather than startingwith a real vector space and trying to create a complexication out of it.

Theorem 3.2. Let V be a nonzero complex vector space and W be a real nonzero subspace.The following are equivalent:

(1) V is the complexication of W ,(2) any R -basis of W is a C -basis of V ,(3) some R -basis of W is a C -basis of V ,(4) the C -linear map C R W V given by z w zw on elementary tensors is an

isomorphism of complex vector spaces,

Proof. (1) (2): Let {e j } be an R -basis of W . Since V = W + iW , V is spanned over Rby {e j , ie j }, so V is spanned over C by {e j }. To show linear independence of {e j } over C ,suppose (a j + ib j )e j = 0. Then a j e j = i ( b j )e j W iW = {0}, so a j = 0 andb j = 0 for all j .

(2) (3): Obvious.


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8 KEITH CONRAD

(3) (4): If some R -basis {e j } of W is a C -basis of V , then the C -linear map C R W V given by z w zw sends 1 e j to e j so it identies C -bases of two complex vectorspaces. Therefore this map is an isomorphism of complex vector spaces.

(4) (1): Since C R W = 1 W + i(1W ) and (1W )i(1W ) = 0, the isomorphism

with V shows V is a complexication of W .By Theorem 3.2, every complex vector space V is the complexication of some real

subspace W : if V = 0, pick a C -basis {e j } of V and take for W the real span of the e j s inV . This is trivial if V = 0.

Example 3.3. Two real subspaces of M 2(C ) having M 2(C ) as their complexication areM2(R ) and {( a b+ cib ci d ) : a,b,c,d R }: both are 4-dimensional real subspaces containinga C -basis of M2(C ) (check!).

Note that saying V has some R -basis of W as a C -basis is stronger than saying V hasa C -spanning set from W , even though a spanning set contains a basis: a spanning setfor V in W means C -coefficients, while a basis of W means R -coefficients, so there is a

mismatch. As an example, let V = W = C . Then V has a C -spanning set in W , e.g. ,{1}, but no R -basis of W is a C -basis of V . In general, a spanning set for V from W neednot be a basis of W . But see Corollary 4.12 for conditions where a spanning set plays animportant role.

Since both W C and C R W have the same properties relative to W insofar as complex-ications are concerned, we can use the label complexication for either construction. Inparticular, when referring to the complexication of an R -linear map : W W , we canmean either C : W C W C or 1 : C R W C R W .

4. Conjugations on Complex Vector Spaces

Real subspaces with a given complexication have a name:

Denition 4.1. If V is a complex vector space, an R - form of V is a real subspace W of V having V as its complexication.

Concretely, an R -form of V is the R -span of any C -basis of V .

Example 4.2. Two R -forms of M2(C ) are M2(R ) and {( a b+ cib ci d ) : a,b,c,d R }.

There is a way to keep track of all the R -forms of a complex vector space by using ageneralization of complex conjugation. When we base extend a real vector space W to thecomplex vector space C R W , complex conjugation can be extended from C to the tensorproduct: let W : C R W C R W by(4.1) W (z w) := z w.This is R -linear, and we can recover W (or rather 1 W inside C R W ) from W by takingxed points. If w W then W (1 w) = 1 w. In the other direction, if W (t) = t forsome t C R W , we want to show t 1W . Write t = 1 w1 + i w2; every elementof C R W has this form in a unique way. Then W (t) = 1 w1 i w2, so W (t) = t if and only if i w2 = i w2, which is the same as 2 i w2 = 0, and that implies w2 = 0,so t = 1 w1 1W . That the real vector space W (really, its isomorphic standard copy1W in C R W ) is the set of xed points of W generalizes the fact that R is the setof xed points of complex conjugation on C : whatever is xed by something like complexconjugation should be thought of as a real object.


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COMPLEXIFICATION 9

Remark 4.3. If we think of the complexication of W as W C = W W , it is natural tocall the complex conjugate of ( w1, w2) = ( w1, 0) + i(w2, 0) the vector ( w1, 0) i(w2, 0) =(w1, w2), so we set (w1, w2) = ( w1, w2). The reader can check the diagram

W Cf W

/ /

v v

C R W W

W C

f W / /C R W

commutes, where f W is the isomorphism in Theorem 3.1. It is easy to see on W C that thexed points of the conjugation are {(w, 0)}, which is the standard copy of W in W C .

Abstracting the function W on C R W to any complex vector space V , we are led tothe following coordinate-free concept.

Denition 4.4. A conjugation on a complex vector space V is a function c: V V suchthat

c(v + v ) = c(v) + c(v ) for all v and v in V , c(zv) = zc(v) for all z C and v V , c(c(v)) = v for all v V .

A conjugation is R -linear (take z R in the second property) but is not C -linear:c(iv) = ic(v). The rst and second properties together are called conjugate-linearity .

Example 4.5. On C n , the function c(z1, . . . , z n ) = ( z1, . . . , z n ), where every coordinategets replaced by its complex conjugate, is a conjugation. In terms of the standard basise1, . . . , e n of C n , this conjugation is given by c( z j e j ) = z j e j . Its xed points are R n .

Example 4.6. On C [X ], the function c( d j =0 a j X j ) = d j =0 a j X

j is a conjugation. Itsxed points are R [X ].

Example 4.7. On M2(C ), two examples of conjugations are c( ) = (

) and c( ) =

( ). The rst one conjugates all the matrix entries, and the second one is the conjugate-transpose operation. The xed points of these conjugations are M 2(R ) and {( a b+ cib ci d ) :a,b,c,d R }, which we met before (Example 3.3) as two R -forms of M2(C ).

Example 4.8. For any real vector space W , the function W in (4.1) is called the standard conjugation on C R W .

Example 4.9. Under the usual C -vector space isomorphisms C R R n = C n and C RR [X ] = C [X ], the standard conjugations on the tensor products are identied with theconjugations on C n and C [X ] from Examples 4.5 and 4.6. That is, the diagrams

C R R n C n

C R R n C n

C R R [X ] C [X ]

C R R [X ] C [X ]

commute, where the horizontal maps are the usual isomorphisms, the left vertical arrowsare the standard conjugations, and the right vertical arrows are the conjugations introducedon C n and C [X ] in Examples 4.5 and 4.6. .


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10 KEITH CONRAD

Example 4.10. If f : V 1 V 2 is an isomorphism of complex vector spaces then a conjuga-tion on V 1 determines a unique conjugation on V 2 such that f transforms the conjugationon V 1 into that on V 2. That is, if the left vertical arrow c1 in the diagram

V 1f

/ /

c1

V 2c2

V 1 f

/ /V 2

is a conjugation on V 1 then there is exactly one conjugation c2 on V 2 that makes the diagramcommute: f (c1(v)) = c2(f (v)) for all v V 1. Indeed, for any v V 2 there is v V 1 suchthat v = f (v). Then c2(v ) = c2(f (v)) = f (c1(v)) = f (c1(f 1(v ))), so we must havec2 = fc 1f 1. Check this formula does dene a conjugation on V 2. In particular, if c is aconjugation on a complex vector space V and A GL(V ) then A is an isomorphism of V with itself and c (v) := Ac(A 1v) is another conjugation on V : the diagram

(4.2) V A /

/c

V c

V A

/ /V

commutes and c is the only conjugation for which that is true.

Example 4.7 reinforces the point that a complex vector space does not have a uniqueconjugation on it. Still, you may feel that the conjugation in Example 4.5 is the mostnatural choice on C n . And you may think R n is the most natural R -form of C n . Thesetwo feelings are related! Here is the link between them, in the most general case:

Theorem 4.11. Let V be a complex vector space. There is a bijection between the following data on V :

(1) R -forms of V ,(2) conjugations on V .

Proof. First we will explain what the maps are in both directions. Then we check the mapsare inverses.

If we start with an R -form W of V , so there is an isomorphism f : C R W V byf (z w) = zw, we can use this isomorphism to transport the standard conjugation W onC R W to a conjugation cW on V (Example 4.10). That is, we can uniquely ll in thearrow on the right of the diagram

C R W f

/ /

W

V cW

C R W

f / /V

to make it commute. Explicitly, cW ( i zi wi ) = i z i wi where zi C and wi W .(Considering the many ways of writing an element of V in the form i zi wi the wi sneed not be a basis the fact that this formula for cW is well-dened depends crucially onC R W being isomorphic to V by z w zw, which is what the two horizontal arrowsare in the diagram.) So from an R -form W of V we get a conjugation cW on V .


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COMPLEXIFICATION 11

If instead we start with a conjugation c on V , an R -form of V is

(4.3) V c = {v V : c(v) = v}.

To check this is an R -form of V , rst note V c is an R -subspace of V since c is R -linear.Then we dene an R -linear map f : C R V c V by f (zw) = zw. We want to show thisis a C -linear isomorphism. Its C -linear since

f (z (z w)) = f (z z w) = ( z z)w = z (zw) = z f (z w),

so f (z t) = z f (t) for all t C R V c by additivity.To check f is onto, we can write any v V as v = w1 + iw2, where w1 = ( v + c(v)) / 2

and w2 = ( v c(v)) / 2i are both xed by c, so each lies in V c. (Compare with the formulas(z + z)/ 2 and (z z)/ 2i for the real and imaginary parts of a complex number z.) Thenv = w1 + iw2 = f (1 w1 + i w2).

To check f is one-to-one, suppose f (t) = 0 for some t C R V c . We can writet = 1 w + i w for some w and w in V c , so the condition f (t) = 0 says w + iw = 0in V . Applying c to this yields w iw = 0 because w and w are xed by c. Adding andsubtracting the equations w + iw = 0 and w iw = 0 shows w = 0 and w = 0, so t = 0.Thus f is one-to-one, so V c is an R -form of V .

It remains to check that our correspondences between R -forms of V and conjugations onV are inverses of one another.

R -form to conjugation and back: If we start with an R -form W of V , we get a conjuga-tion cW on V by cW ( i zi wi ) := i z i wi for zi C and wi W . The subspace V cW is{v V : cW (v) = v}, so we need to check this is W . Certainly W V c .

Since C R W = V by z w zw, every v V is w1 + iw2 for some w1 and w2 in W .Then cW (v) = cW (w1 + iw2) = cW (w1) + cW (iw2) = w1 iw2, so cW (v) = v if and only if iw2 = iw2, which is equivalent to w2 = 0, which means v = w1 W . Thus V cW W , soW = V c.

Conjugation to R -form and back: If we start with a conjugation c on V , so C R V c

= V by zw zw, then we have to check this isomorphism transports the standard conjugation V c on C R V c to the original conjugation c on V (and not some other conjugation on V ). Atensor i ziwi in C R V c is identied with i ziwi in V , and V c ( i ziwi ) = i z iwigoes over to i z i wi in V . Since the wi s are in V c , i z i wi = i z i c(wi ) = c( i zi wi ), sothe isomorphism from C R V c to V does transport V c on C R V c to c on V .

Corollary 4.12. Let V be a complex subspace of C n and c be the usual conjugation on C n .The following conditions are equivalent:

(1) V has a C -spanning set in R n ,(2) c(V ) = V ,(3) V has an R -form in R n ,

(4) dim R (V Rn

) = dim C (V )When this happens, the only R -form of V in R n is V R n .

Proof. This is all clear when V is 0, so take V = 0.(1) (2): If V has a C -spanning set {v j } in R n , c(V ) = c( d j =1 C v j ) =

d j =1 C c(v j ) =

d j =1 C v j = V . Conversely, if c(V ) = V then c is a conjugation on V and the corresponding

R -form of V is V c (C n )c = R n . An R -basis of V c is a C -basis of V , so V has a C -basisin R n .


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12 KEITH CONRAD

(2) (3): In the proof that (2) implies (1) we showed (2) implies (3). In the otherdirection, if (3) holds then there is an R -subspace W R n such that V is the C -span of W . That implies c(V ) = V , since V has a C -spanning in R n .

(1) (4): If V has a C -spanning set in R n then V has a C -basis in R n . Let d = dim C (V ),

and v1, . . . , vd be a C -basis of V in R n . Then every v V has the form v = a j v j witha j C . If v R n then c(v) = v, so a j = a j (from linear independence of the v j s overC ). Therefore v is in the R -span of the v j s, so V R n = R v j has R -dimension d.Conversely, if dim R (R n V ) = dim C (V ), let v1, . . . , vd be an R -basis of V R n . Vectors inR n that are linearly independent over R are linearly independent over C , so the C -span of the v j s has C -dimension d. Since this C -span is in V , whose C -dimension is d, the C -spanof the v j s is V .

It remains to show under these 4 conditions that V R n is the unique R -form of V inR n . By the proof of (4) (1), V R n is an R -form of V . (This also follows from (2) (1)since V c = V (C n )c = V R n .) Now suppose W R n is an R -form of V . Let w1, . . . , wdbe an R -basis of W . The conjugation on V corresponding to the R -form W is given by

a j w j a j w j = c( a j w j ) since w j R n . By Theorem 4.11, W = V R n .

To make Corollary 4.12 concrete, it says in part that the solutions to a homogeneoussystem of complex linear equations is spanned by its real solutions if and only if the solutionspace is preserved by the standard conjugation on C n (in which case there is a homogeneoussystem of real linear equations with the same solutions).

Corollary 4.12 says V R n is an R -form of V only when its R -dimension has the largestconceivable size (since always dim R (V R n ) dim C (V )). It is generally false that V R nspans V over C , since the dimension of the intersection could be too small.

Example 4.13. Let V = C i1 in C2. Then V R 2 = { 00 }. An R -form of this V is R

i1 ,

but there is no R -form of V in R 2.

There is a version of Corollary 4.12 for arbitrary C -vector spaces, not just subspaces of C n or nite-dimensional spaces:

Corollary 4.14. Let U be a complex vector space with a conjugation c on it. For a C -subspace V U , the following conditions are equivalent:

(1) V has a spanning set in U c ,(2) c(V ) = V ,(3) V has an R -form in U c .

When these hold, the only R -form of V in U c is V U c . If dim C (V ) < , these conditions are the same as dim C (V U c) = dim C (V ).

The proof is left as an exercise.As another application of Theorem 4.11, we will prove a converse to the end of Example

4.10: any two conjugations on a complex vector space are conjugate to each other by some(not unique) automorphism of the vector space.


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COMPLEXIFICATION 13

Corollary 4.15. Let V be a complex vector space. For any two conjugations c and c on V , there is some A GL(V ) such that the diagram

(4.4) V A / /

c

V

c

V A / /V

commutes.

Proof. Let W = V c and W = V c . These are both R -forms of V by Theorem 4.11, so thediagrams

C R W / /

W

V c

C R W / /V

C R W / /

W

V

c

C R W / /V

both commute, where the horizontal isomorphisms are given by z w zw and the leftvertical maps are the standard conjugations (Example 4.8).

The real dimensions of W and W are equal (since, as R -forms of V , their real dimensionsboth equal dim C (V )1), so there is an R -linear isomorphism : W W . The base extension1 : C R W CR W is a C -linear isomorphism that respects the standard conjugationon each of these tensor products. That is, the diagram

C R W 1

/ /

W

C R W W

C R W

1

/ /C R W

commutes. (It suffices to check commutativity on elementary tensors: running along thetop and right, W ((1 )(z w)) = W (z (w)) = z (w), and running along theleft and bottom (1 )( W (zw)) = (1 )(z w) = z (w).) Now combine the threediagrams in this proof to get

V / /

c

C R W

W

1 / /C R W

W

/ /V

c

V / /C R W 1 / /C R W / /V.

The maps along the top and bottom are C -linear isomorphisms. Call the (common) com-posite map along the top and bottom A, so A GL(V ), and remove the middle verticalmaps to be left with a commutative diagram of the form (4.4).

Conjugations can be used to describe how the complexications of R -linear maps W 1 W 2 sit inside the C -linear maps C R W 1 C R W 2:

1This could be an innite cardinal number, but what we do is still correct in that case.


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14 KEITH CONRAD

Theorem 4.16. A C -linear map : C R W 1 C R W 2 has the form 1 for some R -linear map : W 1 W 2 if and only if (4.5) ( W 1 (t)) = W 2 ((t))

for all t C R W 1. Here W 1 and W 2 are the standard conjugations on C R W 1 and C R W 2.

Proof. Suppose = 1 for some . To check ( W 1 (t)) = W 2 ((t)) for all t, we only haveto check it on elementary tensors t = zw1: ( W 1 (zw1)) = (1 )(zw1) = z(w1)and W 2 ((z w1)) = W 2 (z (w1)) = z (w1), so (4.5) holds.

Conversely, if satises (4.5) then at t = 1 w1 we get (1 w1) = W 2 ((1 w1)).Thus (1 w1) is xed by the standard conjugation on C R W 2, so (1w1) 1W 2.Every element of 1 W 1 has the form 1 w2 for exactly one w2, so we can dene (w1) W 2by

(1 w1) = 1 ((w1)) .This serves to dene a function : W 1 W 2. Since (w1 + w1) = ( w1) + ( w1),

1 ((w1 + w1)) = 1 ((w1)) + 1 ((w1)) = 1 ((w1) + (w1)) ,so (w1 + w1) = (w1) + (w1). In a similar way one can show commutes with scalingby real numbers, so is R -linear.

Now we want to show = 1 as functions on C R W 1. Both are C -linear, so itsuffices to compare them on elementary tensors of the form 1 w, where

(1 w1) = 1 ((w1)) = (1 )(1 w1).

Theorem 4.16 is related to Theorem 4.11. Dene c: HomC (C R W 1, C R W 2) HomC (C R W 1, C R W 2) by

c() := W 2 W 1 .

Check c is a conjugation on the complex vector space Hom C (C R W 1, C R W 2). Theorem4.16 says c has a naturally associated R -form: { : c() = }. Saying c() = is thesame as W 1 = W 2 (since 1W 2 = W 1 ), which is (4.5). So Theorem 4.16 says theR -form of HomC (C R W 1, C R W 2) corresponding to the conjugation c is {1 : HomR (W 1, W 2)}, which is HomR (W 1, W 2) under the usual identications.

Here is a more general version of Theorem 4.16, starting from complex vector spacesequipped with conjugations.

Theorem 4.17. Let V 1 and V 2 be complex vector spaces equipped with conjugations c1 and c2. A C -linear map : V 1 V 2 has the form 1 for some R -linear map : (V 1)c1 (V 2)c2 if and only if

(c1(t)) = c2((t)) for all t V 1

The proof is left as an exercise.

References

[1] W. Brown, A Second Course in Linear Algebra, Wiley, New York, 1988.[2] P. R. Halmos, Finite-Dimensional Vector Spaces, Springer-Verlag, New York, 1974.[3] I. Satake, Linear Algebra, Marcel Dekker, New York, 1975.

complex if ication

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