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On the other hand, according to Madsen if $4n+2\\neq 2^t-2$ then $M^{4n+2}$ is cobordant to a sphere (hence of Kervaire invariant zero) and $M^{2^{t+1}-2}$ is not cobordant to a sphere (hence of Kervaire invariant one) if and only if certain element $p_{2^{t}-1}^2\\in H_*QS^0$ is spherical. Moreover, it is known t"},"verification_status":{"content_addressed":true,"pith_receipt":true,"author_attested":false,"weak_author_claims":0,"strong_author_claims":0,"externally_anchored":false,"storage_verified":false,"citation_signatures":0,"replication_records":0,"graph_snapshot":true,"references_resolved":false,"formal_links_present":false},"canonical_record":{"source":{"id":"1712.00752","kind":"arxiv","version":2},"metadata":{"license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","primary_cat":"math.AT","submitted_at":"2017-12-03T11:28:56Z","cross_cats_sorted":[],"title_canon_sha256":"0e8fb4f1e91814bb5c65bf2d649d642394aad55769af5edac8e70bbd66629c57","abstract_canon_sha256":"5eac34eb217a460e72ab09c280cd2c76856e16ddad6f905a0eee46984d8791c2"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-18T00:27:17.493370Z","signature_b64":"uTbp4YBFD+z0ZEWxv1Xt6uo0P+mfpSFaMR4ZQ15D8xJvlJuNhD6OfkHExCnkH+ICHBgxDzXzFMlynm6I5u2ZBw==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"7b19ac9e5e72af753f26f85d7f71e542d8f3a338206c8489ec2fa7541048f264","last_reissued_at":"2026-05-18T00:27:17.492869Z","signature_status":"signed_v1","first_computed_at":"2026-05-18T00:27:17.492869Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"Towards a Browder theorem for spherical classes in $\\Omega^lS^{n+l}$","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.AT","authors_text":"Hadi Zare","submitted_at":"2017-12-03T11:28:56Z","abstract_excerpt":"According to Browder if $4n+2\\neq 2^{t+1}-2$ then the Kervaire invariant of the cobordism class of a $(4n+2)$-dimensional manifold $M^{4n+2}$ vanishes and $M^{2^{t+1}-2}$ is of Kervaire invariant one if and only if $h_t^2\\in\\mathrm{Ext}(\\mathbb{Z}/2,\\mathbb{Z}/2)$ is a permanent cycle. On the other hand, according to Madsen if $4n+2\\neq 2^t-2$ then $M^{4n+2}$ is cobordant to a sphere (hence of Kervaire invariant zero) and $M^{2^{t+1}-2}$ is not cobordant to a sphere (hence of Kervaire invariant one) if and only if certain element $p_{2^{t}-1}^2\\in H_*QS^0$ is spherical. 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