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Without that existence, we wouldn't have any reason to believe in even the much weaker consistency assertions. … According to Steel, the large cardinal set-theorist can still consider the nature of V=L set theory, without the stronger large cardinals, simply by relativizing set theory to L. …

What reasons are there to believe in their truth, rather than merely in their consistency, or at most their truth in some transitive model? … One might be $V = L$, or even the axiom of restriction, but most set theorists dismiss them, as they rule out large large cardinals and inaccessible cardinals, respectively, and instead seek core models …

National Acad., 1938, he remarks that V=L added as a new axiom seems to give a natural completion of the axioms of set theory, in so far as it determines the vague notion of an arbitrary infinite set in … So Davis (and apparently also Kreisel) believed that Gödel accepted V=L in 1938 as a new axiom. …

Now, consider the incidence correspondence $X = \lbrace (v,L) \in V \times \mathrm{Gr}(k+1,n+1) \mid v \in L \rbrace \subset V \times \mathrm{Gr}(k+1,n+1)$. This is a projective variety. … \cap L) = 0$ for all $L \in U$. …

Since it is a model of $\text{ZFC}+V=L$, one gets on board all of the consequences of that theory in the minimal model. … But there are many further properties true in the minimal model, when it exists, that are not provable from $\text{ZFC}+V=L$. …

I believe it is essentially this perspective that is expressed by the quotations you have made in the original question. … For example, if ZFC is consistent, then we know that it is consistent that ZFC + V=L + ¬Con(ZFC), with the point being that in any world fulfilling this theory, we would have the full fine-structural account …

way of building up the model $L[A]$), which have to have well-founded ultrapowers (in which case they are called mice), and one needs a comparison lemma for establishing which mice are initial segments … I read some of Steel’s “The Comparison Lemma” in my quest for an answer, and it mentioned that for $L$ the mice are just well-founded structures satisfying a fragment of $\mathrm{ZFC}$ and $V = L$, which …

I don't believe "ultimate L" will ever gain popular traction, for reasons that should be apparent. … I think the question of whether $V$ equals $L$ is far more important to settle. My opinion about $V=L$ is in the minority, but it has had some important supporters. …

There cannot be a "true" $V$. There cannot even be a completed collection of all ordinals. … This rules out wanting $V=L$ as an axiom, as that limits the types of collections available. …

I will assume $V \in L^\infty(\Omega)$ is smooth, $V>0$ and $g \in C^1(\partial \Omega)$. … {L^\infty}}{\lambda_1(\Omega)} \int_\Omega |\nabla u|^2 + 2\|V\|_{L^\infty} \int_\Omega w^2 \,, $$ hence $$ J(u) \ge \left(1-\frac{2\|V\|_{L^\infty}}{\lambda_1(\Omega)} \right) \int_\Omega |\nabla u|^2 …

If we can add one then presumably we can add many, plus if we manage to add one then it already shows the independence of $V=L$ (by absoluteness of $L$ and the fact that $M[G]$ has the same ordinals), … I believe this shows that to some extent, Boolean-valued model does make forcing easier to understand. …

By arithmetical absoluteness, we have $$(\Phi_d^{g''})^{V[G,H]}=(\Phi_d^{g''})^{V[G]}$$ and so $r\in V[G]$; similarly, we get $r\in V[H]$. … On the other hand, if $\mathsf{V=L}$ then there is a $\Delta^1_2$ such $F$ (namely, "Pick the $L$-least upper bound"). …

{(l)}_{i{k_1}} M_{{k_1}k_2}\dots M_{k_{n-1}j} + M_{i{k_1}}V^{(l)}_{{k_1}{k_2}}M_{k_2k_3}\dots M_{k_{n-1}j}+\dots + M_{i{k_1}}M_{k_1k_2}\dots M_{k_{n-2}k_{n-1}}V^{(l)}_{k_{n-1}j}\right)\\ &=\left[V^{(l) … }M^{n-1}\right]_{ij}+\left[M V^{(l)}M^{n-2}\right]_{ij}+\left[M^2V^{(l)}M^{n-3}\right]_{ij}+\dots \left[M^{n-1}V^{(l)}\right]_{ij}\\ &=\sum_{k=0}^{n-1} \left[M^kV^{(l)}M^{n-1-k}\right]_{ij} \end{align} …

The Heisenberg group comes with a natural action of $\Sp(V)$; I'll write it as $h \mapsto h^g$ for $h \in H$ and $g \in \Sp(V)$. … When $k=1$, I believe this is the full normalizer of $H$ in $\SL_p(\CC)$; when $k>1$, I think it is the normalizer of $H$ in $\SL_{p^k}(\QQ(\zeta))$. …

\end{cases} $$ It is easily seen that the variational formulation of the above equation is $$(\nabla u,\nabla v)_{L^2}+(u,v)_{L^2}=(f,v)_{L^2},\;\;\;\;\text{ for all } v\in H^1_{per}(\Omega):=\{v\in H^ … (I believe it's some eigenspace). How to represent solution in this case? …