Collective \( \Gamma \)-Conditions (12)

We clarified the role of the conversion lemma in the rank-g construction last time. It’ll be used to covert a \( \Gamma(b, h) \)-condition given on one-parameter weight \( w_* \) to a \( \Gamma_g \)-condition on the g-parameter weight w primitive with \( w_* \), i.e., w: \( (U_1, U_2, \ldots, U_g ) \mapsto \prod_{i=1}^g w_*(U_i) \). Then we can apply it to \( g^{th} \) mark lemma in order to export the collective \( \Gamma \)-condition.

The lemma on p. 8 is rephrased into:

The Conversion Lemma: if \( \| \cdot \| \) is primitive with \( \| \cdot \|_* \), \[ \Gamma(b, h)~\textrm{of}~ {\mathcal F} ~\textrm{on}~ \| \cdot \|_* ~\textrm{with}~h \ge 1, \qquad \Rightarrow \qquad \Gamma_g \left[ \frac{b}{2^{g-2} (g-1) m} ,~h^{g-1} \right]~\textrm{of}~ {\mathcal F} ~\textrm{on}~ \| \cdot \|. \]

Let’s start its proof. We verify \[ (22) \qquad \| {\mathcal P}_{j, g} \| < h^{g-1} b_g^{-j} {(g-1)m \choose j} \| {\mathcal F} \|^g, \quad \forall j \in [(g-1)m], \quad \textrm{where}~b_g=2^{-g+2} b, \] by induction on \( g \ge 2\) noting that:

  • (22) means the target \( \Gamma_g \)-condition because \( b_g^{-j} {(g-1)m \choose j} \le \left[ \frac{b_g}{(g-1)m} \right]^{-j} \).
  • Paper S4 shows the same for \( g’=[2, g] \). We just confirm it by induction on g here without g’. This simplifies the notation slightly.
  • For a given fixed \( w_* \) inducing \( \| \cdot \|_* \), we show (22) for the \( w \) primitive with \( w_* \), for which we may add the subscript g to write \( w_g \) and \( \| \cdot \|_g \). It’s necessary to distinguish it from the norms for g-1 in the induction step.
  • To distinguish between the original \( \Gamma \) and the \( \Gamma_g \)-condition, we call the former primitive \(\Gamma \)-condition.
  • As said before, the \( \Gamma_g \)-condition must exclude any \( U \not \in {\mathcal F} \) from the norm calculation. It’s achieved by \( U \not \in {\mathcal F} \Rightarrow w_*(U)=0 \) implied by the primitive \( \Gamma \)-condition.

Then prove (22) for the basis \( g=2 \) as follows.

We’ll prove the induction step next time.

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