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			<section>
				<h3>Formalizing</h3>
				<h3> Gardam's Disproof </h3>
				<h5>of</h5>
				<h3> Kaplansky's Unit Conjecture </h3>
				<h5 class="highlight"><NAME></h5>
				<p>Indian Institute of Science</p>
				<h5 class="highlight">  <NAME> </h5>
				<p>IISER, Pune</p>

			</section>


			<section data-transition="zoom-in convex-out" data-background-image="hills.png">
				<h2> Group Rings </h2>
				<h4>and</h4>
				<h2>The Unit Conjecture</h2>
				
			</section>


			<section data-transition="convex" data-background-color="teal">
				<ul>
					<li>Group rings are analogues of polynomials, or better still <em>Laurent polynomials</em>, i.e., finite sums 
					$$\sum_{i=1}^n a_i x^{n_i}.$$</li>
					<li>We take $a_i$ to be elements of a <em>field</em>.</li>
					<li>Observe that, by considering degrees, for two Laurent polynomials $p(x)$ and $q(x)$:
						<ul>
							<li>If $p(x)q(x)=0$, then $p(x)=0$ or $q(x)=0$.</li>
							<li>If $p(x)q(x)=1$, then both $p(x)$ and $q(x)$ are of the form $cx^d$, $c\neq 0$.</li>
						</ul>
					</li>
				</ul>
			</section>

			<section data-transition="convex" data-background-color="teal">
				<h3>Group Rings</h3>
				<ul>
					<li>Fix a group $G$ and a ring $R$.</li>
					<li>The group ring $R[G]$ is the set of formal sums $$\sum_{i=1}^n a_i g_i.$$
						
					</li>
					<li>We multiply bilinearly extending $G$: 
						$$(\sum_{i=1}^k a_i g_i)\cdot (\sum_{j=1}^l b_j h_j) = \sum_{i=1}^k\sum_{j=1}^l (a_ib_j)(g_ih_j).$$
					</li>
				</ul>
			</section>

			<section data-transition="convex" data-background-color="teal">
				<h3>Why Group Rings</h3>
				<ul>
					<li>Any vector space $V$ over $k$ with a linear group of symmetries $G$ forms a <strong> $k[G]$-module</strong>.</li>
					<li>Linear transformations respecting symmetries are given by matrices with entries in the group ring $k[G]$.</li>
					<li>Many fundamental topological invariants, for example, are derived from these matrices.</li>
				</ul>
			</section>

			<section data-transition="convex" data-background-color="teal">
				<ul>
					<li>We consider group rings over fields $k$.</li>
					<li>If an element $g\in G$ is torsion, i.e., $g^n=e$ for some $n>1$, then we have $0$-divisors in $k[G]$ as 
							$$(1-g)(1 + g + \dots + g^{n-1}) = 1 - g^n = 0.$$
					</li>
					<li>
						Similarly, for many groups with torsion, $k[G]$ has <em>non-trivial</em> units, i.e., units not of the form $a\cdot g$.
					</li>
					<li>Kaplansky conjectures: if $G$ is <strong>torsion-free</strong>, $k[G]$ has no zero divisors and no non-trivial units.</li>
					<li>These are related to fundamental questions in topology.</li>
				</ul>
			</section>

			<section data-transition="zoom-in concave-out" data-background-image="campus.jpg">
				<h2 class="highlight">Gardam's Theorem</h2>
				
			</section>

			<section data-background="#b5533c">
				<ul>
					<li><strong>Theorem (Gardam)</strong> Kaplansky's unit conjecture is false.</li>
					<li>Thus,
						<ul>
							<li>For the field $k=\mathbb{F}_2$ with $2$ elements.</li>
							<li>For a certain group $P$, called the <em>Promislow</em> or <em>Hantzsche–Wendt</em> group.</li>
							<li>We have elements $\alpha, \alpha'\in \mathbb{F}_2[P]$ with $\alpha\cdot\alpha'=1$. </li>
							<li>$\alpha$ is not of the form $a\cdot g$.</li>
						</ul>
					</li>
				</ul>
			</section>

			<section data-background="#b5533c">
				<ul>
					<li>The group $P$ is a <em>Metabelian Group</em>, i.e., a group $G$ with an <em>abelian normal subgroup</em> $K$ so that the quotient $Q=G/K$ is also abelian.</li>
					<li>As a set the Metabelian group $G$ is $G = K \times Q$.</li>
					<li>As a group $G$ is determined by 
						<ul>
							<li>The abelian groups $K$ and $Q$.</li>
							<li>A <em>group action</em> of $Q$ on $K$, induced by conjugation in $G$.</li>
							<li>A <em>co-cycle</em> $c: Q \times Q \to Q$.</li>
						</ul>
					</li>
				</ul>
			</section>

			<section data-background="#b5533c">
				<ul>
					<li>
						Torsion-freeness of $P$ follows from:
						<ul><commit_msg>corrections<commit_after><!doctype html>
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	<title>Formalizing Unit Conjecture</title>

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</head>

<body>

	<div class="reveal">

		<!-- Any section element inside of this container is displayed as a slide -->
		<div class="slides">
			<section>
				<h3>Formalizing</h3>
				<h3> Gardam's Disproof </h3>
				<h5>of</h5>
				<h3> Kaplansky's Unit Conjecture </h3>
				<h5 class="highlight"><NAME></h5>
				<p>Indian Institute of Science</p>
				<h5 class="highlight">  <NAME> </h5>
				<p>IISER, Pune</p>

			</section>


			<section data-transition="zoom-in convex-out" data-background-image="hills.png">
				<h2> Group Rings </h2>
				<h4>and</h4>
				<h2>The Unit Conjecture</h2>
				
			</section>


			<section data-transition="convex" data-background-color="teal">
				<ul>
					<li>Group rings are analogues of polynomials, or better still <em>Laurent polynomials</em>, i.e., finite sums 
					$$\sum_{i=1}^n a_i x^{n_i}.$$</li>
					<li>We take $a_i$ to be elements of a <em>field</em>.</li>
					<li>Observe that, by considering degrees, for two Laurent polynomials $p(x)$ and $q(x)$:
						<ul>
							<li>If $p(x)q(x)=0$, then $p(x)=0$ or $q(x)=0$.</li>
							<li>If $p(x)q(x)=1$, then both $p(x)$ and $q(x)$ are of the form $cx^d$, $c\neq 0$.</li>
						</ul>
					</li>
				</ul>
			</section>

			<section data-transition="convex" data-background-color="teal">
				<h4>Group Rings</h4>
				<ul>
					<li>Fix a group $G$ and a ring $R$.</li>
					<li>The group ring $R[G]$ is the set of formal sums $$\sum_{i=1}^n a_i g_i.$$
						
					</li>
					<li>We multiply bilinearly extending $G$: 
						$$(\sum_{i=1}^k a_i g_i)\cdot (\sum_{j=1}^l b_j h_j) = \sum_{i=1}^k\sum_{j=1}^l (a_ib_j)(g_ih_j).$$
					</li>
				</ul>
			</section>

			<section data-transition="convex" data-background-color="teal">
				<h4>Why Group Rings</h4>
				<ul>
					<li>Any vector space $V$ over $k$ with a linear group of symmetries $G$ forms a <strong> $k[G]$-module</strong>.</li>
					<li>Linear transformations respecting symmetries are given by matrices with entries in the group ring $k[G]$.</li>
					<li>Many fundamental topological invariants, for example, are derived from these matrices.</li>
				</ul>
			</section>

			<section data-transition="convex" data-background-color="teal">
				<ul>
					<li>We consider group rings over fields $k$.</li>
					<li>If an element $g\in G$ is torsion, i.e., $g^n=e$ for some $n>1$, then we have $0$-divisors in $k[G]$ as 
							$$(1-g)(1 + g + \dots + g^{n-1}) = 1 - g^n = 0.$$
					</li>
					<li>
						Similarly, for many groups with torsion, $k[G]$ has <em>non-trivial</em> units, i.e., units not of the form $a\cdot g$.
					</li>
					<li>Kaplansky conjectures: if $G$ is <strong>torsion-free</strong>, $k[G]$ has <strong>no zero divisors</strong> and <strong>no non-trivial units</strong>.</li>
					<li>These are also related to fundamental questions in topology, K-theory etc.</li>
				</ul>
			</section>

			<section data-transition="zoom-in concave-out" data-background-image="campus.jpg">
				<h2 class="highlight">Gardam's Theorem</h2>
				
			</section>

			<section data-background="#b5533c">
				<ul>
					<li><strong>Theorem (Gardam)</strong> Kaplansky's unit conjecture is false.</li>
					<li>Thus,
						<ul>
							<li>For the field $k=\mathbb{F}_2$ with $2$ elements.</li>
							<li>For a certain group $P$, called the <em>Promislow</em> or <em>Hantzsche–Wendt</em> group.</li>
							<li>We have elements $\alpha, \alpha'\in \mathbb{F}_2[P]$ with $\alpha\cdot\alpha'=1$. </li>
							<li>$\alpha$ is not of the form $a\cdot g$.</li>
							<li>$P$ is <strong>torsion free.</strong></li>
						</ul>
					</li>
				</ul>
			</section>

			<section data-background="#b5533c">
				<ul>
					<li>The group $P$ is a <em>Metabelian Group</em>, i.e., a group $G$ with an <em>abelian normal subgroup</em> $K$ so that the quotient $Q=G/K$ is also abelian.</li>
					<li>As a set the Metabelian group $G$ is $G = K \times Q$.</li>
					<li>As a group $G$ is determined by 
						<ul>
							<li>The abelian groups $K$ and $Q$.</li>
							<li>A <em>group action</em> of $Q$ on $K$, induced by conjugation in $G$.</li>
							<li>A <em>co-cycle</em> $c: Q \times Q \to Q$.</li>
						</ul>
					</li>
				</ul>
			</section>

			<section data-background="#b5533c">
				<ul>
					<li>
						Torsion-freeness of $P$ follows from:
						<ul>
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