## 45.12 Chern classes

In this section we discuss how given a first Chern class and a projective space bundle formula we can get all Chern classes. A reference for this section is although our axioms are slightly different.

Let $\mathcal{C}$ be a category of schemes with the following properties

1. Every $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ is quasi-compact and quasi-separated.

2. If $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $U \subset X$ is open and closed, then $U \to X$ is a morphism of $\mathcal{C}$. If $X' \to X$ is a morphism of $\mathcal{C}$ factoring through $U$, then $X' \to U$ is a morphism of $\mathcal{C}$.

3. If $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and if $\mathcal{E}$ is a finite locally free $\mathcal{O}_ X$-module, then

1. $p : \mathbf{P}(\mathcal{E}) \to X$ is a morphism of $\mathcal{C}$,

2. for a morphism $f : X' \to X$ in $\mathcal{C}$ the induced morphism $\mathbf{P}(f^*\mathcal{E}) \to \mathbf{P}(\mathcal{E})$ is a morphism of $\mathcal{C}$,

3. if $\mathcal{E} \to \mathcal{F}$ is a surjection onto another finite locally free $\mathcal{O}_ X$-module then the closed immersion $\mathbf{P}(\mathcal{F}) \to \mathbf{P}(\mathcal{E})$ is a morphism of $\mathcal{C}$.

Next, assume given a contravariant functor $A$ from the category $\mathcal{C}$ to the category of graded algebras. Here a graded algebra $A$ is a unital, associative, not necessarily commutative $\mathbf{Z}$-algebra $A$ endowed with a grading $A = \bigoplus _{i \geq 0} A^ i$. Given a morphism $f : X' \to X$ of $\mathcal{C}$ we denote $f^* : A(X) \to A(X')$ the induced algebra map. We will denote the product of $a, b \in A(X)$ by $a \cup b$. Finally, we assume given for every object $X$ of $\mathcal{C}$ an additive map

$c_1^ A : \mathop{\mathrm{Pic}}\nolimits (X) \longrightarrow A^1(X)$

We assume the following axioms are satisfied

1. Given $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $\mathcal{L} \in \mathop{\mathrm{Pic}}\nolimits (X)$ the element $c_1^ A(\mathcal{L})$ is in the center of the algebra $A(X)$.

2. If $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $X = U \amalg V$ with $U$ and $V$ open and closed, then $A(X) = A(U) \times A(V)$ via the induced maps $A(X) \to A(U)$ and $A(X) \to A(V)$.

3. If $f : X' \to X$ is a morphism of $\mathcal{C}$ and $\mathcal{L}$ is an invertible $\mathcal{O}_ X$-module, then $f^*c_1^ A(\mathcal{L}) = c_1^ A(f^*\mathcal{L})$.

4. Given $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and locally free $\mathcal{O}_ X$-module $\mathcal{E}$ of constant rank $r$ consider the morphism $p : P = \mathbf{P}(\mathcal{E}) \to X$ of $\mathcal{C}$. Then the map

$\bigoplus \nolimits _{i = 0, \ldots , r - 1} A(X) \longrightarrow A(P),\quad (a_0, \ldots , a_{r - 1}) \longmapsto \sum c_1^ A(\mathcal{O}_ P(1))^ i \cup p^*(a_ i)$

is bijective.

5. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and let $\mathcal{E} \to \mathcal{F}$ be a surjection of finite locally free $\mathcal{O}_ X$-modules of ranks $r + 1$ and $r$. Denote $i : P' = \mathbf{P}(\mathcal{F}) \to \mathbf{P}(\mathcal{E}) = P$ the corresponding incusion morphism. This is a morphism of $\mathcal{C}$ which exhibits $P'$ as an effective Cartier divisor on $P$. Then for $a \in A(P)$ with $i^*a = 0$ we have $a \cup c_1^ A(\mathcal{O}_ P(P')) = 0$.

To formulate our result recall that $\textit{Vect}(X)$ denotes the (exact) category of finite locally free $\mathcal{O}_ X$-modules. In Derived Categories of Schemes, Section 36.38 we have defined the zeroth $K$-group $K_0(\textit{Vect}(X))$ of this category. Moreover, we have seen that $K_0(\textit{Vect}(X))$ is a ring, see Derived Categories of Schemes, Remark 36.38.6.

Proposition 45.12.1. In the situation above there is a unique rule which assigns to every $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ a “total Chern class”

$c^ A : K_0(\textit{Vect}(X)) \longrightarrow \prod \nolimits _{i \geq 0} A^ i(X)$

with the following properties

1. For $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we have $c^ A(\alpha + \beta ) = c^ A(\alpha ) c^ A(\beta )$ and $c^ A(0) = 1$.

2. If $f : X' \to X$ is a morphism of $\mathcal{C}$, then $f^* \circ c^ A = c^ A \circ f^*$.

3. Given $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $\mathcal{L} \in \mathop{\mathrm{Pic}}\nolimits (X)$ we have $c^ A([\mathcal{L}]) = 1 + c_1^ A(\mathcal{L})$.

Proof. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module. We first show how to define an element $c^ A(\mathcal{E}) \in A(X)$.

As a first step, let $X = \bigcup X_ r$ be the decomposition into open and closed subschemes such that $\mathcal{E}|_{X_ r}$ has constant rank $r$. Since $X$ is quasi-compact, this decomposition is finite. Hence $A(X) = \prod A(X_ r)$. Thus it suffices to define $c^ A(\mathcal{E})$ when $\mathcal{E}$ has constant rank $r$. In this case let $p : P \to X$ be the projective bundle of $\mathcal{E}$. We can uniquely define elements $c_ i^ A(\mathcal{E}) \in A^ i(X)$ for $i \geq 0$ such that $c_0^ A(\mathcal{E}) = 1$ and the equation

45.12.1.1
$$\label{weil-equation-chern-classes} \sum \nolimits _{i = 0}^ r (-1)^ i c_1(\mathcal{O}_ P(1))^ i \cup p^*c^ A_{r - i}(\mathcal{E}) = 0$$

is true. As usual we set $c^ A(\mathcal{E}) = c_0^ A(\mathcal{E}) + c_1^ A(\mathcal{E}) + \ldots + c_ r^ A(\mathcal{E})$ in $A(X)$.

If $\mathcal{E}$ is invertible, then $c^ A(\mathcal{E}) = 1 + c_1^ A(\mathcal{L})$. This follows immediately from the construction above.

The elements $c_ i^ A(\mathcal{E})$ are in the center of $A(X)$. Namely, to prove this we may assume $\mathcal{E}$ has constant rank $r$. Let $p : P \to X$ be the corresponding projective bundle. if $a \in A(X)$ then $p^*a \cup (-1)^ r c_1(\mathcal{O}_ P(1))^ r = (-1)^ r c_1(\mathcal{O}_ P(1))^ r \cup p^*a$ and hence we must have the same for all the other terms in the expression defining $c_ i^ A(\mathcal{E})$ as well and we conclude.

If $f : X' \to X$ is a morphism of $\mathcal{C}$, then $f^*c_ i^ A(\mathcal{E}) = c_ i^ A(f^*\mathcal{E})$. Namely, to prove this we may assume $\mathcal{E}$ has constant rank $r$. Let $p : P \to X$ and $p' : P' \to X'$ be the projective bundles corresponding to $\mathcal{E}$ and $f^*\mathcal{E}$. The induced morphism $g : P' \to P$ is a morphism of $\mathcal{C}$. The pullback by $g$ of the equality defining $c_ i^ A(\mathcal{E})$ is the corresponding equation for $f^*\mathcal{E}$ and we conclude.

Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Consider a short exact sequence

$0 \to \mathcal{L} \to \mathcal{E} \to \mathcal{F} \to 0$

of finite locally free $\mathcal{O}_ X$-modules with $\mathcal{L}$ invertible. Then

$c^ A(\mathcal{E}) = c^ A(\mathcal{L}) c^ A(\mathcal{F})$

Namely, by the construction of $c^ A_ i$ we may assume $\mathcal{E}$ has constant rank $r + 1$ and $\mathcal{F}$ has constant rank $r$. The inclusion

$i : P' = \mathbf{P}(\mathcal{F}) \longrightarrow \mathbf{P}(\mathcal{E}) = P$

is a morphism of $\mathcal{C}$ and it is the zero scheme of a regular section of the invertible module $\mathcal{L}^{\otimes -1} \otimes \mathcal{O}_ P(1)$. The element

$\sum \nolimits _{i = 0}^ r (-1)^ i c_1^ A(\mathcal{O}_ P(1))^ i \cup p^*c^ A_ i(\mathcal{F})$

pulls back to zero on $P'$ by definition. Hence we see that

$\left(c_1^ A(\mathcal{O}_ P(1)) - c_1^ A(\mathcal{L})\right) \cup \left(\sum \nolimits _{i = 0}^ r (-1)^ i c_1^ A(\mathcal{O}_ P(1))^ i \cup p^*c^ A_ i(\mathcal{F})\right) = 0$

in $A^*(P)$ by assumption (5) on our cohomology $A$. By definition of $c_1^ A(\mathcal{E})$ this gives the desired equality.

Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Consider a short exact sequence

$0 \to \mathcal{E} \to \mathcal{F} \to \mathcal{G} \to 0$

of finite locally free $\mathcal{O}_ X$-modules. Then

$c^ A(\mathcal{F}) = c^ A(\mathcal{E}) c^ A(\mathcal{G})$

Namely, by the construction of $c^ A_ i$ we may assume $\mathcal{E}$, $\mathcal{F}$, and $\mathcal{G}$ have constant ranks $r$, $s$, and $t$. We prove it by induction on $r$. The case $r = 1$ was done above. If $r > 1$, then it suffices to check this after pulling back by the morphism $\mathbf{P}(\mathcal{E}^\vee ) \to X$. Thus we may assume we have an invertible submodule $\mathcal{L} \subset \mathcal{E}$ such that both $\mathcal{E}' = \mathcal{E}/\mathcal{L}$ and $\mathcal{F}' = \mathcal{E}/\mathcal{L}$ are finite locally free (of ranks $s - 1$ and $t - 1$). Then we have

$c^ A(\mathcal{E}) = c^ A(\mathcal{L}) c^ A(\mathcal{E}') \quad \text{and}\quad c^ A(\mathcal{F}) = c^ A(\mathcal{L}) c^ A(\mathcal{F}')$

Since we have the short exact sequence

$0 \to \mathcal{E}' \to \mathcal{F}' \to \mathcal{G} \to 0$

we see by induction hypothesis that

$c^ A(\mathcal{F}') = c^ A(\mathcal{E}') c^ A(\mathcal{G})$

Thus the result follows from a formal calculation.

At this point for $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we can define $c^ A : K_0(\textit{Vect}(X)) \to A(X)$. Namely, we send a generator $[\mathcal{E}]$ to $c^ A(\mathcal{E})$ and we extend multiplicatively. Thus for example $c^ A(-[\mathcal{E}]) = c^ A(\mathcal{E})^{-1}$ is the formal inverse of $a^ A([\mathcal{E}])$. The multiplicativity in short exact sequences shown above guarantees that this works.

Uniqueness. Suppose $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $\mathcal{E}$ is a finite locally free $\mathcal{O}_ X$-module. We want to show that conditions (1), (2), and (3) of the lemma uniquely determine $c^ A([\mathcal{E}])$. To prove this we may assume $\mathcal{E}$ has constant rank $r$; this already uses (2). Then we may use induction on $r$. If $r = 1$, then uniqueness follows from (3). If $r > 1$ we pullback using (2) to the projective bundle $p : P \to X$ and we see that we may assume we have a short exact sequence $0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{E}'' \to 0$ with $\mathcal{E}'$ and $\mathcal{E}''$ having lower rank. By induction hypothesis $c^ A(\mathcal{E}')$ and $c^ A(\mathcal{E}'')$ are uniquely determined. Thus uniqueness for $\mathcal{E}$ by the axiom (1). $\square$

Lemma 45.12.2. In the situation above. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Let $\mathcal{E}_ i$ be a finite collection of locally free $\mathcal{O}_ X$-modules of rank $r_ i$. There exists a morphism $p : P \to X$ in $\mathcal{C}$ such that

1. $p^* : A(X) \to A(P)$ is injective,

2. each $p^*\mathcal{E}_ i$ has a filtration whose successive quotients $\mathcal{L}_{i, 1}, \ldots , \mathcal{L}_{i, r_ i}$ are invertible $\mathcal{O}_ P$-modules.

Proof. We may assume $r_ i \geq 1$ for all $i$. We will prove the lemma by induction on $\sum (r_ i - 1)$. If this integer is $0$, then $\mathcal{E}_ i$ is invertible for all $i$ and we conclude by taking $\pi = \text{id}_ X$. If not, then we can pick an $i$ such that $r_ i > 1$ and consider the projective bundle $p : P \to X$ associated to $\mathcal{E}_ i$. We have a short exact sequence

$0 \to \mathcal{F} \to p^*\mathcal{E}_ i \to \mathcal{O}_ P(1) \to 0$

of finite locally free $\mathcal{O}_ P$-modules of ranks $r_ i - 1$, $r_ i$, and $1$. Observe that $p^* : A(X) \to A(P)$ is injective by assumption. By the induction hypothesis applied to the finite locally free $\mathcal{O}_ P$-modules $\mathcal{F}$ and $p^*\mathcal{E}_{i'}$ for $i' \not= i$, we find a morphism $p' : P' \to P$ with properties stated as in the lemma. Then the composition $p \circ p' : P' \to X$ does the job. $\square$

Lemma 45.12.3. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Then

$c^ A_ i({\mathcal E} \otimes {\mathcal L}) = \sum \nolimits _{j = 0}^ i \binom {r - i + j}{j} c^ A_{i - j}({\mathcal E}) \cup c^ A_1({\mathcal L})^ j$

Proof. By the construction of $c^ A_ i$ we may assume $\mathcal{E}$ has constant rank $r$. Let $p : P \to X$ and $p' : P' \to X$ be the projective bundle associated to $\mathcal{E}$ and $\mathcal{E} \otimes \mathcal{L}$. Then there is an isomorphism $g : P \to P'$ such that $g^*\mathcal{O}_{P'}(1) = \mathcal{O}_ P(1) \otimes p^*\mathcal{L}$. See Constructions, Lemma 27.20.1. Thus

$g^*c_1^ A(\mathcal{O}_{P'}(1)) = c_1^ A(\mathcal{O}_ P(1)) + p^*c_1^ A(\mathcal{L})$

The desired equality follows formally from this and the definition of Chern classes using equation (45.12.1.1). $\square$

Proposition 45.12.4. In the situation above assume $A(X)$ is a $\mathbf{Q}$-algebra for all $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Then there is a unique rule which assigns to every $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ a “chern character”

$ch^ A : K_0(\textit{Vect}(X)) \longrightarrow \prod \nolimits _{i \geq 0} A^ i(X)$

with the following properties

1. $ch^ A$ is a ring map for all $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

2. If $f : X' \to X$ is a morphism of $\mathcal{C}$, then $f^* \circ ch^ A = ch^ A \circ f^*$.

3. Given $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $\mathcal{L} \in \mathop{\mathrm{Pic}}\nolimits (X)$ we have $ch^ A([\mathcal{L}]) = \exp (c_1^ A(\mathcal{L}))$.

Proof. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module. We first show how to define the rank $r^ A(\mathcal{E}) \in A^0(X)$. Namely, let $X = \bigcup X_ r$ be the decomposition into open and closed subschemes such that $\mathcal{E}|_{X_ r}$ has constant rank $r$. Since $X$ is quasi-compact, this decomposition is finite, say $X = X_0 \amalg X_1 \amalg \ldots \amalg X_ n$. Then $A(X) = A(X_0) \times A(X_1) \times \ldots \times A(X_ n)$. Thus we can define $r^ A(\mathcal{E}) = (0, 1, \ldots , n) \in A^0(X)$.

Let $P_ p(c_1, \ldots , c_ p)$ be the polynomials constructed in Chow Homology, Example 42.43.6. Then we can define

$ch^ A(\mathcal{E}) = r^ A(\mathcal{E}) + \sum \nolimits _{i \geq 1} (1/i!) P_ i(c^ A_1(\mathcal{E}), \ldots , c^ A_ i(\mathcal{E})) \in \prod \nolimits _{i \geq 0} A^ i(X)$

where $ci^ A$ are the Chern classes of Proposition 45.12.1. It follows immediately that we have property (2) and (3) of the lemma.

We still have to show the following three statements

1. If $0 \to \mathcal{E}_1 \to \mathcal{E} \to \mathcal{E}_2 \to 0$ is a short exact sequence of finite locally free $\mathcal{O}_ X$-modules on $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, then $ch^ A(\mathcal{E}) = ch^ A(\mathcal{E}_1) + ch^ A(\mathcal{E}_2)$.

2. If $\mathcal{E}_1$ and $\mathcal{E}_2 \to 0$ are finite locally free $\mathcal{O}_ X$-modules on $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, then $ch^ A(\mathcal{E}_1 \otimes \mathcal{E}_2) = ch^ A(\mathcal{E}_1) ch^ A(\mathcal{E}_2)$.

Namely, the first will prove that $ch^ A$ factors through $K_0(\textit{Vect}(X))$ and the first and the second will combined show that $ch^ A$ is a ring map.

To prove these statements we can reduce to the case where $\mathcal{E}_1$ and $\mathcal{E}_2$ have constant ranks $r_1$ and $r_2$. In this case the equalities in $A^0(X)$ are immediate. To prove the equalities in higher degrees, by Lemma 45.12.2 we may assume that $\mathcal{E}_1$ and $\mathcal{E}_2$ have filtrations whose graded pieces are invertible modules $\mathcal{L}_{1, j}$, $j = 1, \ldots , r_1$ and $\mathcal{L}_{2, j}$, $j = 1, \ldots , r_2$. Using the multiplicativity of Chern classes we get

$c_ i^ A(\mathcal{E}_1) = s_ i(c_1^ A(\mathcal{L}_{1, 1}), \ldots , c_1^ A(\mathcal{L}_{1, r_1}))$

where $s_ i$ is the $i$th elementary symmetric function as in Chow Homology, Example 42.43.6. Similarly for $c_ i^ A(\mathcal{E}_2)$. In case (1) we get

$c_ i^ A(\mathcal{E}) = s_ i(c_1^ A(\mathcal{L}_{1, 1}), \ldots , c_1^ A(\mathcal{L}_{1, r_1}), c_1^ A(\mathcal{L}_{2, 1}), \ldots , c_1^ A(\mathcal{L}_{2, r_2}))$

and for case (2) we get

$c_ i^ A(\mathcal{E}_1 \otimes \mathcal{E}_2) = s_ i(c_1^ A(\mathcal{L}_{1, 1}) + c_1^ A(\mathcal{L}_{2, 1}), \ldots , c_1^ A(\mathcal{L}_{1, r_1}) + c_1^ A(\mathcal{L}_{2, r_2}))$

By the definition of the polynomials $P_ i$ we see that this means

$P_ i(c^ A_1(\mathcal{E}_1), \ldots , c^ A_ i(\mathcal{E}_1)) = \sum \nolimits _{j = 1, \ldots , r_1} c_1^ A(\mathcal{L}_{1, j})^ i$

and similarly for $\mathcal{E}_2$. In case (1) we have also

$P_ i(c^ A_1(\mathcal{E}), \ldots , c^ A_ i(\mathcal{E})) = \sum \nolimits _{j = 1, \ldots , r_1} c_1^ A(\mathcal{L}_{1, j})^ i + \sum \nolimits _{j = 1, \ldots , r_2} c_1^ A(\mathcal{L}_{2, j})^ i$

In case (2) we get accordingly

$P_ i(c^ A_1(\mathcal{E}_1 \otimes \mathcal{E}_2), \ldots , c^ A_ i(\mathcal{E}_1 \otimes \mathcal{E}_2)) = \sum \nolimits _{j = 1, \ldots , r_1} \sum \nolimits _{j' = 1, \ldots , r_2} (c_1^ A(\mathcal{L}_{1, j}) + c_1^ A(\mathcal{L}_{2, j'}))^ i$

Thus the desired equalities are now consequences of elementary identities between symmetric polynomials.

We omit the proof of uniqueness. $\square$

Lemma 45.12.5. In the situation above let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. If $\psi ^2$ is as in Chow Homology, Lemma 42.56.1 and $c^ A$ and $ch^ A$ are as in Propositions 45.12.1 and 45.12.4 then we have $c^ A_ i(\psi ^2(\alpha )) = 2^ i c^ A_ i(\alpha )$ and $ch^ A_ i(\psi ^2(\alpha )) = 2^ i ch^ A_ i(\alpha )$ for all $\alpha \in K_0(\textit{Vect}(X))$.

Proof. Observe that the map $\prod _{i \geq 0} A^ i(X) \to \prod _{i \geq 0} A^ i(X)$ multiplying by $2^ i$ on $A^ i(X)$ is a ring map. Hence, since $\psi ^2$ is also a ring map, it suffices to prove the formulas for additive generators of $K_0(\textit{Vect}(X))$. Thus we may assume $\alpha = [\mathcal{E}]$ for some finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$. By construction of the Chern classes of $\mathcal{E}$ we immediately reduce to the case where $\mathcal{E}$ has constant rank $r$. In this case, we can choose a projective smooth morphism $p : P \to X$ such that restriction $A^*(X) \to A^*(P)$ is injective and such that $p^*\mathcal{E}$ has a finite filtration whose graded parts are invertible $\mathcal{O}_ P$-modules $\mathcal{L}_ j$, see Lemma 45.12.2. Then $[p^*\mathcal{E}] = \sum [\mathcal{L}_ j]$ and hence $\psi ^2([p^*\mathcal{E}]) = \sum [\mathcal{L}_ j^{\otimes 2}]$ by definition of $\psi ^2$. Setting $x_ j = c^ A_1(\mathcal{L}_ j)$ we have

$c^ A(\alpha ) = \prod (1 + x_ j) \quad \text{and}\quad c^ A(\psi ^2(\alpha )) = \prod (1 + 2 x_ j)$

in $\prod A^ i(P)$ and we have

$ch^ A(\alpha ) = \sum \exp (x_ j) \quad \text{and}\quad ch^ A(\psi ^2(\alpha )) = \sum \exp (2 x_ j)$

in $\prod A^ i(P)$. From these formulas the desired result follows. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).