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Here the characteristic curves are straight lines $x=y+sa$, $y\cdot n=0$, along which $\dot{u}=c\cdot (y+sa)+d$. If $a\cdot n\ne 0$, every nearby point determines a unique starting point and travel parameter:Integrating along the Characteristic curve givesThe non-characteristic condition is exactly what permits division by $a\cdot n$ and ensures transversality to the data hyperplane. If $a\cdot n=0$, the Characteristic curve lines starting there never leave it. Moreover the prescribed data must satisfy $a\cdot {\nabla }_S{u}_0=c\cdot x+d$ on the hyperplane even to be compatible. If they do, the Characteristic curve construction still leaves off-hyperplane data undetermined; if they do not, no $C^1$ solution exists.

For a first-order quasilinear partial differential equation, take $C^1$ coefficients $a_j,b$, $C^1$ prescribed data $u_0$ on a smooth $C^2$ hypersurface $S$, and a point $x_0\in S$. The data are non-characteristic there ifwhere $n$ is any nonzero normal to $S$. Then there is a neighborhood of $x_0$ on which a unique $C^1$ solution realizes the data. Uniqueness is local, after restricting to a neighborhood swept out by the nearby characteristics.

The method of characteristics solves $\dot{x}=a(x,u)$, $\dot{u}=b(x,u)$ from points of $S$. In local surface coordinates $\eta$, the map $(\eta ,s)\mapsto x(\eta ,s)$ has nonsingular derivative at $s=0$ exactly under the stated normal condition. The inverse function theorem makes these variables local coordinates, giving the asserted solution and uniqueness.

The Black-Scholes implied volatility of a traded call price $C$ is the volatility $\sigma$ which makes the Black-Scholes formula equal to $C$. Forthe price is a continuous strictly increasing function of $\sigma >0$, by its positive vega, with precisely these endpoint limits. The intermediate value theorem and strict monotonicity therefore give a unique positive implied volatility. The lower endpoint corresponds to the limiting value $\sigma =0$; the upper endpoint is reached only as $\sigma \to \infty$. Prices outside the bounds admit the arbitrages above and have no implied volatility.

Write $\varphi ={\Phi }^{\prime }$ for the standard normal density. Differentiating the Black-Scholes formula and using $S_0\varphi (d_1)=K{e}^{-rT}\varphi (d_2)$ givesThe last equality follows by differentiating $d_1-d_2=\sigma \sqrt{T}$. This derivative is the option's vega.

As $\sigma \downarrow 0$, the Black-Scholes formula tends to $\boxed{(S_0-K{e}^{-rT}{)}^{+}}$. This also follows by discounting the deterministic zero-volatility payoff; at equality $S_0=K{e}^{-rT}$ the limiting value is zero.

If $S_0>K{e}^{-rT}$ and a call trades at $C<S_0-K{e}^{-rT}$, buy a call, short one share, and buy a bond paying $K$ at maturity. Invest the positive initial surplus $S_0-K{e}^{-rT}-C$. At maturity the call, share and bond together pay $(S_T-K{)}^{+}-S_T+K=(K-S_T{)}^{+}\ge 0$, in addition to the strictly positive invested surplus. If the lower bound is zero, violating it means $C<0$: buy the call and invest the cash received, obtaining a nonnegative option payoff and a strictly positive bank payoff. Both strategies are arbitrages.

As $\sigma \to \infty$, $d_1\to +\infty$ and $d_2\to -\infty$, so the European call option price tends to $\boxed{S_0}$. If its traded price $C>S_0$, sell a call, buy one share, and invest the surplus $C-S_0$ at the risk-free rate. The initial net cost is zero, and the maturity payoff isThis is an explicit arbitrage.

For positive $S_0,K,T,\sigma$, the Black-Scholes formula usesTo hedge the sold European call option, buy $\Phi (d_1)$ shares and borrow $K{e}^{-rT}\Phi (d_2)$ in cash. This long replicating strategy has initial value $S_0\Phi (d_1)-K{e}^{-rT}\Phi (d_2)$, financed by the option premium. The share holding must subsequently be rebalanced according to the option's option delta in a self-financing strategy.

Write $q_{ij}=c{m}_{ij}$ for $i\ne j$, with a common positive constant $c$, and put $d_i={\sum }_{j\ne i}{m}_{ij}$. The infinitesimal generator matrix is symmetric because the graph is undirected. Thus the uniform distribution ${\pi }_i=1/|I|$ satisfies detailed balance, and the continuous-time random walk is reversible when started uniformly.

Its jump chain chooses a link uniformly from those incident to its current vertex:The flux ${\hat{\pi }}_i{\hat{P}}_{ij}=m_{ij}/{\sum }_j{d}_j$ is symmetric, so this random walk on a multigraph is also reversible. The continuous-time equilibrium is uniform because higher-degree vertices have shorter holding times; at jump times the equilibrium is degree-weighted. If the intended proportionality constant depends on the starting vertex, $q_{ij}=c_i{m}_{ij}$, reversibility still holds with ${\pi }_i\propto 1/c_i$, and the same jump chain results.

Use the equilibrium correspondence from the preceding part. For $i\ne j$,Consequently the detailed balance equations for the jump chain hold exactly when those for the continuous-time Markov chain hold. The reversibility characterizations already given prove the equivalence. Both chains here are started in their respective stationary distributions, which generally differ.

The jump chain records the successive states visited at actual jumps of the continuous-time Markov chain. Put $q_i=-q_{ii}$. For an irreducible chain with at least two states, all $q_i>0$, and its transition probabilities are ${\hat{P}}_{ij}=q_{ij}/q_i$ for $j\ne i$, with ${\hat{P}}_{ii}=0$. The holding time in state $i$ is exponential with mean $1/q_i$.

If $\pi Q=0$, set $C={\sum }_i{\pi }_i{q}_i$ and ${\hat{\pi }}_i={\pi }_i{q}_i/C$. ThenConversely, $\hat{\pi }\hat{P}=\hat{\pi }$ implies that ${\pi }_i=c{\hat{\pi }}_i/q_i$, normalized to sum one, satisfies $\pi Q=0$. ThereforeJump observations overrepresent states with short holding times. A one-state chain has no actual jumps; its trivial equilibrium can be handled separately, without using this division by $q_i=0$.

A Discrete-time Markov chain with initial distribution $\hat{\pi }$ is reversible if $(Y_0,\dots ,Y_n)$ and $(Y_n,\dots ,Y_0)$ have the same law for every $n$. Its detailed balance equations are ${\hat{\pi }}_i{\hat{P}}_{ij}={\hat{\pi }}_j{\hat{P}}_{ji}$. These equations are equivalent to reversibility. Indeed, reversing every factor in the probability ${\hat{\pi }}_{i_0}{\prod }_{r=1}^n{\hat{P}}_{i_{r-1}{i}_r}$ gives the reverse path probability; the converse follows by taking $n=1$.

Summing detailed balance over $i$ givesHence $\hat{\pi }$ is an stationary distribution.

Let $\pi$ be a probability vector satisfying detailed balance. Since each row of the infinitesimal generator matrix sums to zero,Thus $\pi Q=0$, and $\pi e^{tQ}=\pi$ for all $t\ge 0$. This proves that $\pi$ is an stationary distribution. The diagonal term causes no problem because its detailed balance equation is an identity.

A continuous-time Markov chain started with distribution $\pi$ is reversible if its finite-dimensional distributions are unchanged by reversing time: for every $T>0$, $(X_t:0\le t\le T)$ and $(X_{T-t}:0\le t\le T)$ have the same law. In particular, reversibility includes stationarity. The detailed balance equations areFor a finite continuous-time Markov chain, these equations are equivalent to reversibility with initial distribution $\pi$. Equivalently, they imply ${\pi }_i{P}_{ij}(t)={\pi }_j{P}_{ji}(t)$ for every transition matrix $P(t)=e^{tQ}$.

For $n=1$, the singular set is the point $\{0\}$, a zero-dimensional submanifold. For $n\ge 2$, every elementary matrix $E_{ij}$ is singular and the smooth curve $t\mapsto t{E}_{ij}$ lies in $X$ through zero. If $X$ were a submanifold, all $E_{ij}$ would belong to $T_{0}X$ and span $M(n)$. Its dimension at zero would therefore be $n^2$, making $X$ locally open there. But arbitrarily small matrices $tI$ are invertible, a contradiction. ThusThe failure is at the zero matrix; fixed-rank pieces of the singular set can separately be smooth submanifolds.

For an invertible $A$, factor $\det (A+tB)=\det A\det (I+t{A}^{-1}B)$. The determinant's permutation expansion gives $\det (I+tC)=1+t\mathrm{tr}C+O(t^2)$. Differentiating at zero yieldsAt a determinant-one matrix choose $B=A/n$; the derivative is one. Thus one is a regular value of the smooth determinant map on $GL(n)$. The regular level set theorem makes $SL(n)$ a submanifold of codimension of a submanifold one, withThis includes $n=1$, when $SL(1)=\{1\}$ has dimension zero.

The determinant is continuous and $GL(n)={\det }^{-1}(\mathbb{R}\setminus \{0\})$ is open in $M(n)\simeq {\mathbb{R}}^{n^2}$. Every open subset inherits the ambient smooth charts, so it is a submanifold and

A geometric circle is an unknot. Compactify ${\mathbb{R}}^3$ to $S^3$, and carry the circle to $\{(z_1,z_2)\in {\mathbb{C}}^2:|z_1{|}^2+|z_2{|}^2=1,z_2=0\}$. Its complement has coordinates $z_1\in \Delta$ and $z_2=\sqrt{1-|z_1{|}^2}{e}^{i\theta }$, making it homeomorphic to $\Delta \times S^1$. It deformation retracts to $S^1$. Removing the additional point at infinity does not change the fundamental group: van Kampen for a small ball about that point uses a simply connected punctured ball, homotopy equivalent to $S^2$. ThusA meridian linking the circle once generates it. This uses “circle” in its geometric sense; the complement of an arbitrary knotted embedded copy of $S^1$ need not have this group.

Choose open neighborhoods of the sphere and circle in the wedge, each enlarged by a small contractible neighborhood of the attaching point in the other piece. They deformation retraction to $S^2$ and $S^1$ and have contractible intersection. Van Kampen therefore givesThe generator traverses the circle once; the sphere contributes no fundamental-group generator.

The van Kampen theorem says: if $X=U\cup V$ with $U,V$ open and path connected, $U\cap V$ path connected, and a base point in the intersection, then ${\pi }_1(X)$ is the free product ${\pi }_1(U)\ast {\pi }_1(V)$ modulo the normal subgroup generated by identifying the two images of every element of ${\pi }_1(U\cap V)$. Equivalently it is the pushout of the two inclusion-induced fundamental-group maps.

The element $\omega$ satisfies ${\omega }^2-\omega +6=0$, so is an algebraic integer. Conversely, write an algebraic integer as $(a+b\sqrt{-23})/2$ with $a,b$ rational. Its trace gives $a\in \mathbb{Z}$, and its field norm implies $23b^2\in \mathbb{Z}$. If $b=p/q$ in lowest terms, $q^2$ divides 23, forcing $q=1$. The integral field norm condition modulo four then makes $a,b$ have the same parity. Therefore it equals $(a-b)/2+b\omega$, provingBoth reductions of $x^2-x+6$ modulo two and three split into the distinct roots zero and one. The index is one, so ideal factorization givesTheir norms are two and three respectively. The product ${\mathfrak{p}}_2{\mathfrak{p}}_3$ is generated by $6,2\omega ,3\omega ,{\omega }^2$. All are multiples of $\omega$, using $6=\omega \overline{\omega }$; and $\omega =3\omega -2\omega$ belongs to the product. HenceThe Minkowski bound is $(2/\pi )\sqrt{23}<4$, in fact about $3.05$. Every class therefore has a representative of ideal norm one, two or three. The displayed primes and their conjugates exhaust those possibilities. Their classes are all $1,c,c^{-1}$, with $c=[{\mathfrak{p}}_2]$, because $[{\mathfrak{p}}_3]=c^{-1}$ and conjugation inverts a quadratic ideal class. There are at most three classes.

There is no element of field norm two: $N(a+b\omega )=a^2+ab+6b^2=(a+b/2{)}^2+23b^2/4$, which exceeds two for $b\ne 0$ and is a square for $b=0$. Thus ${\mathfrak{p}}_2$ is not principal. Moreover $N(\omega -2)=8$, and $\omega -2$ belongs to ${\mathfrak{p}}_2$ but not its conjugate, so $(\omega -2)={\mathfrak{p}}_2^3$. Consequently $c$ has order three, and