Exercises 4.3 Lie algebra A vector space G is called a Lie algebra if there exists a product (called Lie bracket) [
;
] : G  G ! G with the following properties

 Antisymmetry: [x; y] = ????[y; x] ; 8x; y 2 G

 Jacobi identity: [x; [y; z]] + [z; [x; y]] + [y; [z; x]] = 0 ; 8x; y; z 2 G 1. Prove that the space of real antisymmetric n-dimensional matrix U(n) = fA 2 GL(n;R) ; Ay = ????Ag is a (nite) Lie algebra with the bracket

[A;B] = AB ???? BA 2. Prove that the space of vector elds of TM is an (innite) Lie algebra with the Lie bracket

[X; Y ] = (Xi@iY j ???? Y i@iXj)@j where we have set X = Xi@i and Y = Y i@i.

Exercises 4.4 Asian option and Heisenberg algebra An Asian call option on a single stock with strike K is dened by the following payo at the maturity date T max

1 T

Z T 0

fudu ???? K; 0

!

with fu the stock forward at time u (we take a zero interest rate for the sake of simplicity). We explain in this exercise a link with the Heisenberg Lie algebra.

We note At =

R t 0 fudu. The SDEs generated by the processes (ft;At) in the Black-Scholes model are dft ft

= dWt dAt = ftdt 1. Convert the It^o diusion in a Stratonovich diusion dft ft = ????
1 22dt +   dWt dAt = ftdt 2. Write the diusion operator in the Hormander form.
3. Convert these SDEs into ODEs by replacing the Brownian Wt by the path ! = t.
dft ft = (????
1 22  )dt dAt = ftdt 4. These ows (see B.6 in appendix B) are generated by the following vector elds V = f@A + (????
1 22  )f@f Prove that the vector elds V+, V????, V+????  [V+; V????] satisfy the following Lie algebra (see exercise 4.3), called Heisenberg algebra [V+????; V+] = 0 ; [V+????; V????] = 0 Exercises 4.5 Mehler's formula In this exercise, we give an example of a one-dimensional heat kernel equation (HKE) which admits an analytical solution. We consider the following HKE @t p(t; xjx0) = (@2 x ????
r2x2 16 ???? f)p(t; xjx0) (4.85)
with r and f two constants.
1. By scaling the variables x and t and doing a time-dependent gauge transform, show that the HKE (4.85) can be reduced to @tp(t; xjx0) = (@2 x ???? x2)p(t; xjx0)
This PDE corresponds to a HKE with a zero Abelian connection A and a quadratic potential Q (called harmonic potential). By observing that the operator D is quadratic in dierentiation and multiplication, we seek a solution which is a Gaussian function of x and x0. In addition, the solution must clearly be symmetric in x and x0 since the operator D is self-adjoint. We therefore try the following ansatz p(t; xjx0) = exp 
a(t)x2 2 + b(t)xx0 + a(t)x20 + c(t)

2. Prove that the coecients a(t); b(t) and c(t) satisfy the following ODEs _a(t)
2 = a(t)2 ???? 1 = b(t)2 _ c(t) = a(t)
3. Prove that the solutions are given by a(t) = ????coth(2t + C)
b(t) = csc(2t + C)
c(t) = ????
1 2 ln sinh(2t + C) + D with C and D two integration constants.
4. By using the initial condition p(0; xjx0) = (x ???? x0), show that the values of the integration constants are C = 0 D = ln(2)????1 2 5. Finally, prove that the solution of (4.85) is p(t; xjx0) = s t r 2 4t sinh(t r 2 )
exp 
????
r 8t 
coth(t r 2 )(x2 + x20 ) ???? 2 csc(t r 2 )xx0 
???? tf