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\documentclass{article}
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\usepackage[utf8]{inputenc}
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\usepackage{amsfonts}
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\usepackage{amsthm}
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\usepackage{amsmath}
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\usepackage{enumerate}
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\usepackage{hyperref}
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colorlinks,
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\usepackage{xcolor}
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% prevent warnings of underfull \hbox:
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\usepackage{etoolbox}
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\apptocmd{\sloppy}{\hbadness 4000\relax}{}{}
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\theoremstyle{definition}
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\newtheorem{definition}{Def}[section]
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\newtheorem{theorem}[definition]{Thm}
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\title{Notes on BLS Signatures}
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\author{arnaucube}
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\date{July 2022}
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\begin{document}
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\maketitle
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\begin{abstract}
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Notes taken while reading about BLS signatures \cite{bls-sig-eth2}. Usually while reading papers I take handwritten notes, this document contains some of them re-written to $LaTeX$.
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The notes are not complete, don't include all the steps neither all the proofs.
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\end{abstract}
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% \tableofcontents
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\section{BLS signatures}
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\paragraph{Key generation}
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$sk \in \mathbb{Z}_q$, $pk = [sk] \cdot g_1$, where $g_1 \in G_1$, and is the generator.
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\paragraph{Signature}
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$$\sigma = [sk] \cdot H(m)$$
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where $H$ is a function that maps to a point in $G_2$. So $H(m), \sigma \in G_2$.
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\paragraph{Verification}
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$$e(g_1, \sigma) == e(pk, H(m))$$
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Unfold:
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$$e(pk, H(m)) = e([sk] \cdot g_1, H(m) = e(g_1, H(m))^{sk} = e(g_1, [sk] \cdot H(m)) = e(g_1, \sigma))$$
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\paragraph{Aggregation}
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Signatures aggregation:
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$$\sigma_{aggr} = \sigma_1 + \sigma_2 + \ldots + \sigma_n$$
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where $\sigma_{aggr} \in G_2$, and an aggregated signatures is indistinguishible from a non-aggregated signature.
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\vspace{0.5cm}
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Public keys aggregation:
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$$pk_{aggr} = pk_1 + pk_2 + \ldots + pk_n$$
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where $pk_{aggr} \in G_1$, and an aggregated public keys is indistinguishible from a non-aggregated public key.
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\paragraph{Verification of aggregated signatures}
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Identical to verification of a normal signature as long as we use the same corresponding aggregated public key:
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$$e(g_1, \sigma_{aggr})==e(pk_{aggr}, H(m))$$
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Unfold:
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$$e(pk_{aggr}, H(m))= e(pk_1 + pk_2 + \ldots + pk_n, H(m)) =$$
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$$=e([sk_1] \cdot g_1 + [sk_2] \cdot g_1 + \ldots + [sk_n] \cdot g_1, H(m))=$$
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$$=e([sk_1 + sk_2 + \ldots + sk_n] \cdot g_1, H(m))=$$
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$$=[sk_1 + sk_2 + \ldots + sk_n]~\cdot~e(g_1, H(m))=$$
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$$=e(g_1, [sk_1 + sk_2 + \ldots + sk_n] \cdot H(m))=$$
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$$=e(g_1, [sk_1] \cdot H(m) + [sk_2] \cdot H(m) + \ldots + [sk_n] \cdot H(m))=$$
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$$=e(g_1, \sigma_1 + \sigma_2 + \ldots + \sigma_n)= e(g_1, \sigma_{aggr})$$
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Note: in the current notes $pk \in G_1$ and $\sigma, H(m) \in G_2$, but we could use $\sigma, H(m) \in G_1$ and $pk \in G_2$.
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\bibliography{paper-notes.bib}
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\bibliographystyle{unsrt}
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\end{document}
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