{"id":118,"date":"2019-01-08T00:51:15","date_gmt":"2019-01-08T00:51:15","guid":{"rendered":"https:\/\/petermc.net\/blog\/?p=118"},"modified":"2023-10-14T01:28:36","modified_gmt":"2023-10-14T01:28:36","slug":"an-exceptional-isomorphism","status":"publish","type":"post","link":"https:\/\/petermc.net\/blog\/2019\/01\/08\/an-exceptional-isomorphism\/","title":{"rendered":"An exceptional isomorphism"},"content":{"rendered":"

We will construct the exceptional isomorphism \"S_6\cong Sp_4(\mathbb{F}_2)\".<\/p>\n

The group \"S_6\" acts on \"\mathbb{F}_2^6\" preserving the usual pairing \"\langle e_i,e_j\rangle=\delta_{ij}\" where the \"e_i\" are the usual basis vectors.<\/p>\n

There is an invariant line \"L\", the span of \"\sum_i e_i\" and an invariant hyperplane \"H=\{\sum_i a_ie_i|\sum_i a_i=0\}\". Let \"V=H/L\". \"S_6\" acts on \"V\".<\/p>\n

Since \"L\" is the radical of the pairing \"\langle \cdot,\cdot\rangle\" on \"H\", the pairing \"\langle \cdot,\cdot\rangle\" descends to a non-degenerate bilinear pairing on \"V\". As it is symmetric and we are in characteristic 2, it is automatically skew-symmetric.<\/p>\n

The \"S_6\"-action preserves this pairing, hence we get our desired homomorphism from \"S_6\" to \"Sp_4(\mathbb{F}_2)\".<\/p>\n

To check injectivity, it suffices to show that \"(12)\" is not in the kernel, since we know all normal subgroups of \"S_6\". Surjectivity then follows by a counting argument, so we get our desired isomorphism.<\/p>\n","protected":false},"excerpt":{"rendered":"

We will construct the exceptional isomorphism . The group acts on preserving the usual pairing where the are the usual basis vectors. There is an invariant line , the span of and an invariant hyperplane . Let . acts on … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[6],"tags":[],"class_list":["post-118","post","type-post","status-publish","format-standard","hentry","category-maths"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":false,"jetpack_shortlink":"https:\/\/wp.me\/p7V6a7-1U","_links":{"self":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts\/118","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/comments?post=118"}],"version-history":[{"count":7,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts\/118\/revisions"}],"predecessor-version":[{"id":382,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts\/118\/revisions\/382"}],"wp:attachment":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/media?parent=118"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/categories?post=118"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/tags?post=118"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}