samples/texts/1749790/page_14.md

Now, we can pass to the limit as $\lambda \to \infty$: If

$$B=\underset{\lambda \to \mathrm{\infty }}{\lim }{B}_{\lambda }^{-1},B\; =\; \backslash lim\_\{\backslash lambda\; \backslash to\; \backslash infty\}\; B\_\{\backslash lambda\}^\{-1\},$$

the limiting algebra has the structure

Further, if $B_{\lambda}^{-1}$ is analytic $Y_{\lambda}$ in the neighborhood of infinite, then the formulas (6.1) show that the algebra for large $\lambda$ is a perfectly smooth deformation in the Kodaira-Spencer sense of the $\infty$-algebra, which we denote by $G_{\infty}$.

The structure of $G_{\infty}$ can be exhibited quite nicely if $B$ is a projection operator $B^2 = B$ as it is for the case where $G(Q)$ is the Poincaré group, and $G_{\infty}$ is the Galilean group. (There, $B_{\lambda}$ is the diagonal matrix

$\lambda = c^2; c = \text{velocity of light}$ and $B$ is the matrix

Then

$$\begin{array}{cccc}& T=BT\oplus (I-B)T,Q(BT,(I-B)T)=Q(T,B(I-B)T)=0B^2=B,\text{ and }B=B^{\ast }\text{).}& & \text{(since )}\end{array}T\; =\; BT\; \backslash oplus\; (I\; -\; B)T,\; \backslash \backslash \; Q(BT,\; (I\; -\; B)T)\; =\; Q(T,\; B(I\; -\; B)T)\; =\; 0\; \backslash tag\{since\; \}\; B^2\; =\; B,\; \backslash text\{\; and\; \}\; B\; =\; B^*\; \backslash text\{).\}$$

Let

$$A=I-2B\mathrm{.}A\; =\; I\; -\; 2B.$$

Then, $s^2 = I + 4B^2 - 4B = I$.

$$Q(sX,sY)=Q(X,s^{2}Y)=Q(X,Y)\mathrm{.}Q(sX,\; sY)\; =\; Q(X,\; s^2Y)\; =\; Q(X,\; Y).$$

Thus, $s$ is an automorphism of $T$ whose square is the identity which preserves the form $Q: s$ defines a symmetric automorphism of $K(Q)$ by the formula:

$$s(A)=sAs\text{for}A\in K(Q)\mathrm{.}s(A)\; =\; sAs\; \backslash quad\; \backslash text\{for\}\; \backslash quad\; A\; \backslash in\; K(Q).$$

Let $L$ be the set of all $A \in K(Q)$ such that

$$s(A)=A\mathrm{.}s(A)\; =\; A.$$

Let $P$ be the set of all $A \in K(Q)$ such that

$$s(A)=A\mathrm{.}s(A)\; =\; A.$$