Computing the Grothendieck-Springer resolution for $G = SL_2$ Let $G = SL_2, \mathfrak{g} = \mathfrak{sl}_2$, $B$ the Borel subgroup, and $\mathfrak{u}$ the unipotent radical; so that $G/B = \mathbb{P}^1$; how does $\widetilde{\mathfrak{g}}$ decompose as a vector bundle over $\mathbb{P}^1$? Recall the definition:
$\widetilde{\mathfrak{g}} = $ {$(X, gB) \in \mathfrak{g}^* \times \mathbb{P}^1 | X|_{g \mathfrak{u}} = 0$}
This should be simple, but I'm having trouble. 
 A: A paper of Bezrukavnikov (http://arxiv.org/abs/math/0604445) identifies this as $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$ in Example 2.8.  What follows is an argument.
The identification of $G$-equivariant vector bundles on $G/B$ with $B$-representations gives us a short exact sequence of $B$-representations:
$$0 \rightarrow V_{-2} \rightarrow \mathfrak{b} \rightarrow V_0 \rightarrow 0$$
which gives us a short exact sequence in vector bundles:
$$0 \rightarrow \mathcal{O}(-2) \rightarrow \tilde{\mathfrak{sl}_2} \rightarrow \mathcal{O}(0) \rightarrow 0$$
There is no decomposition theorem of $SL_2$-equivariant vector bundles on $\mathbb{P}^1$, since this category is equivalent to $B$-representations and $B$ is solvable.  I claim that if we base change to the non-equivariant case, then the short exact sequence becomes:
$$0 \rightarrow \mathcal{O}(-2) \rightarrow \mathcal{O}(-1) \oplus \mathcal{O}(-1) \rightarrow \mathcal{O}(0) \rightarrow 0$$
Let $\mathcal{E}$ be the locally free sheaf on $\mathbb{P}^1$ associated to $\tilde{\mathfrak{sl}_2}$.  Take the long exact sequence of the short exact sequence of locally free sheaves and one obtains:
$$0 \rightarrow H^0(\mathbb{P}^1, \mathcal{E}) \rightarrow H^0(\mathbb{P}^1, \mathcal{O}(0)) \rightarrow H^1(\mathbb{P}^1, \mathcal{O}(-2)) \rightarrow H^1(\mathbb{P}^1, \mathcal{E}) \rightarrow 0.$$
I claim that the first map must be zero.  By Borel-Weil-Bott all cohomologies $H^i(\mathbb{P}^1, \mathcal{E})$ are sums of trivial representations (since $H^0(\mathbb{P}^1, \mathcal{O}(0) = H^1(\mathbb{P}^1, \mathcal{O}(-2))$ are trivial); in particular all global sections are $G$-invariant.  Above the Borel $$\left(\begin{array}{cc} * & * \\ 0 & * \end{array}\right)$$
take the point in the fiber
$$\left(\begin{array}{cc} a & b \\ 0 & -a \end{array}\right)$$
and use the $G$-action to try to move it around.  In particular, the action of
$$\left(\begin{array}{cc} 1 & t \\ 0 & 1 \end{array}\right)$$
fixes the Borel but moves the section to
$$\left(\begin{array}{cc} a & -2at + b \\ 0 & -a \end{array}\right)$$
so the only section which has a chance of invariant under $G$ has $a = 0$ on this fiber (we haven't even considered whether it extends globally).  But on this fiber it maps to zero to $\mathcal{O}(0)$ and so by equivariance, if it extends to a global section, this section also maps to zero.  So the first map in the short exact sequence is zero, thus the second map is an isomorphism, and the third map is zero.  So $\mathcal{E}$ has no cohomology, and by preservation of rank, it must be $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$.
Note: this bundle is as a "twist" of the usual short exact sequence for the tautological bundle on $\mathbb{P}^1$:
$$0 \rightarrow \mathcal{O}(-1) \rightarrow \mathcal{O}(0) \oplus \mathcal{O}(0) \rightarrow \mathcal{O}(1) \rightarrow 0.$$
A: Let $\frak{b}\subseteq\frak{g}$ denote the Lie algebra of your Borel $B$. There is a natural $G$-equivariant isomorphism $\tilde{\frak{g}}\cong G\times_B\frak{b}$ of vector bundles over $G/B$, where $G\times_B\frak{b}$ is the associated bundle arising from the adjoint representation of $B$ on $\frak{b}$. Let us take $B$ to be the standard Borel of upper-triangular matrices of unit determinant, and consider the maximal torus $T\subseteq B$ of diagonal matrices. Under the adjoint representation restricted to $T$, $\frak{b}$ decomposes into irreducible subrepresentations of weights $0$ and $2$. Conjecturally, anyway, I would not be surprised if you could use this to show that $\tilde{\frak{g}}\cong\mathcal{O}(0)\oplus\mathcal{O}(2)$. Still, I am not sure about this, and there are a few details to be resolved. I hope this is useful.
A: Hi Vinoth, here are my thoughts, hopefully they're correct and what you're after: 
You can think of $\mathfrak{\tilde{g}}$ as the set of pairs
$$\{(X,L)\in \mathfrak{g}\times \mathbb{P}^{1}\;|\; X(L)\subset L\}$$
(identify $\mathfrak{g}$ and $\mathfrak{g}^{\ast}$ via the Killing form). Thus, the fibre over $L\in \mathbb{P}^{1}$ is the the set of $X\in \mathfrak{sl}_{2}$ that have $L$ as an 'eigenline'. Since the projection to $G/B$ is $G$-equivariant it suffices to consider a particular line $L_{0}$; take $L_{0}=span\{e_{1}\}$, to see that the fibre is the standard Borel $\mathfrak{b}$, so that the fibre over $g\cdot L_{0}$ is $g\cdot\mathfrak{b}$.
EDIT: providing the answer in terms of the dual $\mathfrak{g}^{\ast}$, as the question states, we see that the fibre is the span of the dual basis (via the Killing form) of the standard basis $\{F,H\}$ of the lower triangular Borel.
