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Let everything be over $\mathbb{C}$. Consider two varieties $X,$ $Y,$ where $X$ is normal and $Y$ is affine, having regular $\mathbb{C}^*$-actions and a $\mathbb{C}^*$-equivariant projective morphism $\pi: X \rightarrow Y.$ The "projective" means that it factors trough an closed immersion and projection, $$\pi=\pi_Y \circ \iota : X \hookrightarrow \mathbb{P}^n \times Y \rightarrow Y$$ for some $n.$

Now, the question: Can this immersion be made $\mathbb{C}^*$-equivariant? That is, having a linear $\mathbb{C}^*$-action on $\mathbb{P} ^ M$ for some $M$ such that $\iota$ is $\mathbb{C}^*$-equivariant, where on $\mathbb{P} ^M \times Y$ we consider the diagonal action.

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  • $\begingroup$ I think if X is normal, then locally at least Sumihiro’s Theorem should come into play. May I ask what the intended application is? $\endgroup$
    – rvk
    Commented Nov 2, 2021 at 21:12

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No. Let $Y$ be a point and $X$ be $\mathbb P^1$ with two points glued together. Let $G=\mathbb G_m$ act on $X$ by its usual action on $\mathbb P^1$ fixing those two points, and acting trivially (of course) on $Y$.

Then this morphism is not equivariantly projective because, for any $\mathbb G_m$ action on $\mathbb P^M$, and any $x \in \mathbb P^M$ not fixed, the two limits of the $\mathbb G_m$-translates of $x$ as the parameter in $\mathbb G_m$ goes to $0$ and $\infty$ must be distinct.

One can check this by working in coordinates which are eigenvectors for the $\mathbb G_m$ action - one limit involves taking the nonvanishing coordinates of $x$ where $\mathbb G_m$ acts by the greatest power and one by taking the nonvanishing coordinates of $x$ where $\mathbb G_m$ acts by the least power, and these are only equal if $\mathbb G_m$ acts by only a single power on nonvanishing coordinates of $x$, in which case $x$ is fixed by $\mathbb G_m$.

Conversely, for any smooth point on $x$, the two limits of the $\mathbb G_m$-translates are the same, nodal, point.

So you will at least need to assume normalcy or something on $X$ and not just $Y$.

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  • $\begingroup$ Thanks, Will! That explains that $X$ better be smooth -- I added that condition, which is actually the case in which I am interested. $\endgroup$
    – Filip
    Commented Nov 3, 2021 at 9:33
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    $\begingroup$ @Filip92 Asking whether there is a $\mathbb{G}_m$-equivariant projective embedding seems to be equivalent to asking whether there is a $\pi$-relatively ample line bundle on $X$ that is $\mathbb{G}_m$-equivariant. If you assume that $X$ is normal, then you can take any $\pi$-relatively ample line bundle $\mathcal{L}$ on $X$ and then it is known that some power $\mathcal{L}^{\otimes n}$ will admit a $\mathbb{G}_m$-linearization. See e.g. Theorem 2.14 in arxiv.org/pdf/1312.6267.pdf. Alternatively, I think that this is also proved in Mumford's GIT book. $\endgroup$
    – afh
    Commented Nov 3, 2021 at 22:48
  • $\begingroup$ Thanks, @afh this now makes sense then! I might write a full answer in detail, in case you don't want to. $\endgroup$
    – Filip
    Commented Nov 5, 2021 at 0:48
  • $\begingroup$ If you want to write the missing details in a full answer you are welcome to. Probably the main thing to note is that $Y$ being affine allows you to get a graded surjection from a free graded $\mathcal{O}_{Y}$-module into the global sections of $\mathcal{L}^{\otimes n}$ (where $n$ is chosen large enough). $\endgroup$
    – afh
    Commented Nov 5, 2021 at 2:06
  • $\begingroup$ (Regardless, I think that Will Sawin's answer should probably be accepted, since he correctly addressed the original version of the question before edits.) $\endgroup$
    – afh
    Commented Nov 5, 2021 at 2:20

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