Different Geometry Mini case, Day 3

-

That : If filmN"

, } swoth & ofEN is a rgww " he off

,

then ftlq ) is a smooth submitted of M f diners in m - n.

pony th , } ksif boils dam to write this in nice teal cordials & apply the hnsefmhh theorem.

NAI.

The case is not the, as you can see by cnsidriy fi 5 '

→ R gin by flay,z)=zt

.

0, } a rrithl Wu of f

, yet

ft 1o) = the quetr B a smooth I - dink submitted.

by.. Dye f ; Rn → IR by fl × , . -

,xD = Exit .

Then all r > 0 are regw rates,

so the sphere f ' '( r ) f radius rr , }

a smooth sub mnifld of diners in nt.

i Chat.

Use the theorem to shtht the ottogalgpp OH = { A an nxnmtic : A A's= ID B a manifold

.

let Matnxyk ) be the set f real nxn metrics,

which can be identified with RY

In turn,let Symmln ) be the set of nxn symmetric mtias

,uhdn can be identified with A"÷ Define

fi Matnalk → Symmln ) by

A M AAT

Then 04 = f ' ( ID ,the inwxingef the identity metix, so the chdbp B to shot

dfq : Ta Matnxnlk) → TI Symmln ) B sujectk for all AE OH.

To prove th}.

ntetht TA Mata.IR) an be idnlilndwl Metnnlk ) & Tpfymmk ) on be ikdtrd Y synuk ).

Then,for A E f YI ) & CETISKNH

,

I claim tht dfalk CA ) = C.

to see this,let

a : H,4 → Nhtnnlk ) st. a lot = A & a

' lo ) =HCA. Then

dfa KCA) = dfqx ' lo ) = ( fox )'

6) = adthafdlt ) NAT ) = 210) a' lost +2%12 lost

=

AHAy + t CA At

= k AATCT + ECAAT

= kct t t (

= ( since AAT = In & C symmelm.

thefe,04 = till In ) B .

smooth mi told of diners in n'

. HI! nh÷l .

A similar gnat stars tht UH,

the sit f nxn unity minus,

} an ni dinesh manifold.

Nnthtwehmetayut reeks,

man also talk abut vector fields :

Df.A reefer field X on a smooth manifold M is a smooth sdkflte target bullet M

, many a smooth

map × : M →TM so tht A oh PCM),

Hp) ETPM .

Note : Tf4 : UEIR "→ M B a coordinate chat in a nhd f p,

then XIP) = Zailp ) Fxi,nhu each ai : an → R

B swath & { Ex:} B the ksis of Tpm associated to I U, 6).

. A new field X an M ab as a mp CTM ) → Mm ) by

Hf) Ip) = Eailp ) ¥Flp in toil cards.

by : - V = - ykxtx he an 1122- \

⇐tie-

anq 1

he: V = . xtx . 'By 't ' - Asean 5 €*\ g9

a r s

Its a theorem 1add the Hairy Ball Theory tht no smooth ( or em ok no , ) rear field on 5 can be mvnishiy

everywhere.

at: Aside 5 as the wit qhetrnins : 53 = { atsitcjtde : a 2+5 ti tat B

Then fr pe 5,

we on think of Tps3

as the qmtrnhs nhthmpoprdialr top .Far exgde

,1Es3&T±s3={ aitbjtck}

Nn,

Ian define 3 mntlforthonrlmelvfdds X,Y,Z on 5 as Hbo :

Far ah pEs3, X (p) = ip ,Hp) =jp ,

Ztp) = kp .

Ayhtbtthseaetapt to p : If p= pot p,it pzjt Bk ,

then

4 p, ip) = ( pot P,it pzjt Bk ,

- p, tpoi - Bit Rk ) = -

pop, tp , Po- RB +13 R = 0

,ek

.

And you an s.mil cheek tat they're orthomoml at oh point.

So this B striking different from the 5 sky . . .

here we be an eh 4,3 of ham ishg met fields a 53,

Df: lit X be a uf. on M

.

A ane at - E,E) → M is a integral one fr × if a

'H = XKHH fall tete ,{ )

.

next12 the example

,the irtepd any of the votr field X a 5 are exactly the ants in the Hpffbrthskn an Monday

,

In coadhts,the ktgnl are Tzhp B 5pH) = last tisnt ) pteitz, ,

eitzD= eittz,, z , ) uhez

,= pot ip ,

,z ,

= pstipg,

So you a see th } B the ihtbeek f the aphx the tfh p Hz , a) Et w / the Sphe .

In psklr, by albpsy each 5pA) to a point,

we get . mp 5→9P '⇒

'called the Hpf mp .

but: Consider QD,the mnitld of orthogal metros from take

.

Then an dent ftp.QD is a tagat vat,r i.e., theft

to a are d : t - SE ) → an ) st . al o) = I.

Then alt ) EQD F t, so we knew AHKHT = I H t

.Now different ite :

0 = tattle.( at Halts ) = 21012%5+2 ' lo) a lost Fa

'lost to' 6)

,

Since 210) = I = 21 of

so he see tht 2'lo ) = - a

' lost , so d' 'D B a Shen - symaetn matrix

.

Thebe,

he an think f TIN

as the space f nxn Sheu - symmetn minus.

Now,fer QE oh )

,wht does Tuan ) look like ?

let La : 04 → Oh) be sm byLalA) = QA.

then (dLa)[D= QD ( cheek this if ya oh't believe it ! )

Sine 1dL

a),

: TIOH → To,01D B tdrk

,its swjedne & so ay dat f TAOH G be with as

1dL

atA = QD fr sane Isetan )

.

Here,

an arbityuyfr field X on Oh) an be repented by X (Q) = Q Da,uhe Da , } skw - symuelm .

In tam, if Da B oytt with ant to Q 1 mean if

× ' 9 = Qb for se fixed d),then the

new field X , a QD is an exapkf a left - involved field anthe Lie grp 04, ad

Also,

the veetr tells X, Yit on 53 are exgdyf right . hwiatveetr fields on the Lie grp

8.

Lieb

Df: A Lie grip 6 is a group whih is also adifferentiable mnifld so tht the mp 6×6 → 6 gin by ↳ , Htgh

"

B smooth.

(As a argue ,mukptuth & thy mess a both smooth)

by: . 53,

She the pnshtf to wit quieten is is a wit qutmin : Hpyl = Hp" Kyl .

. 013),

She the pooht f othool retries Barthol .

. Mae gulf , ay mtiogrps : GUN,R )

,GUN

,E)

,SUHR)

,SLK

,e)

,OH

,Sold

,Uk)

,Suk )

. The try Th = (syn = (Ulm

. The Hamby grp { 1 !! } ) }. RIP base RP3± So b) (why ?)

Df : Left & r '3ht talk an a lie gp G ae dying y , frg EG, Lglhtgh & BK) = hg , respeelmf .

A vgtr field X on 6 is called left - innit if dLgX=X VGEG & Balled right . that if dRgX=X FGEG.

hj .For the v. f. X an 53 gin by Hpk ip,

notice tht

(dRg)p×#ldRDp? where Spltk last tistlp

=lRq°Dko)=¥l⇐d5pHq)=F¥⇐.last fist ) Pg

= ipq = X ( pp,

so X B r .3ht . Kwit.

§ '

.

For

Xdm0tDsmbyXdlQ5QD.htiithffwAEdny@LnDaXb1Ql-ldLaDo.Q

A = AQ D= XIA a),

so X,

B left . in wit.

Ingenet,

a left - Wright tmwint rear field X Bopbtf detunedby its rhe Heft He ihttg ef 6 !

Tf X is LI & GEG,

then Xlg ) =(dLg)eXk ) .

Caverly ,if v ETEG

,then neon define a left - muwit vat fed X by XD = CdLD ev .

So we can identify the coHalm f kfthwit rear fields an 6 (ok called the Lie algebra f 6) with the

tamgt space to He identity TEG.

We saw befetht TIQD osisb f the nxnshw - symmetric mtnb,

so the left ' ihmnwt new Adds an 04 are in

bijutm anespmbe w/the Shu . sym ah nxn menus,

with the bisecting in hg

betty - X. .

.