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Let $\kappa>0$ and $d,k$ be a positive integers with $k\ge d$. For $k\in \mathbb{Z}^+$ large enough, can one find a geodesically complete and simply connected $d$-dimensional Riemannian submanifold $(M_{\kappa},g)$ of the $k$-fold product of the hyperbolic plane $\prod_{i=1}^k\, \mathbb{H}^2$ with sectional curvature bounded in $[-\kappa,0)$?

Ideally, can we take it to have constant sectional curvature $-\kappa$?

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    $\begingroup$ Just use Nash embedding theorem with target a $k$-flat in the product. $\endgroup$ Commented Jan 7, 2023 at 23:23
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    $\begingroup$ @MoisheKohan I'm guessing you mean $d$-flat? Though this does ofc answer the first question, it won't work for the $-\kappa$-sectionally curbed submanifold. $\endgroup$ Commented Jan 7, 2023 at 23:40
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    $\begingroup$ No, I meant exactly what I wrote. $\endgroup$ Commented Jan 7, 2023 at 23:42
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    $\begingroup$ @MoisheKohan But this has dimension $k$ and not $d$; unless you mean we set $d=k$? $\endgroup$ Commented Jan 7, 2023 at 23:45
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    $\begingroup$ Maybe we use different definitions of Riemannian submanifolds. Please, edit your question to clarify. $\endgroup$ Commented Jan 7, 2023 at 23:48

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For the $i$th factor ${\mathbb H}^2$ in the product of hyperbolic planes, pick a complete geodesic $c_i$, $i=1,...,k$. The product $$ F=c_1\times ... \times c_k\subset X=\prod_{i=1}^k {\mathbb H}^2 $$ is a $k$-flat, i.e. a totally-geodesic (although we do not need this) isometrically embedded Euclidean subspace of dimension $k$. Let $(M,g)$ be a $d$-dimensional Riemannian manifold (no restrictions whatsoever). By the Nash isometric embedding theorem, there exists an isometric embedding $$ f: (M,g)\to F, $$ provided that $k$ is much larger than $d$. (Specifically, you can take any $k\ge (3d+11)/2$.) Then the composition of $f$ with the identity embedding $F\to X$, gives an isometric embedding $(M,g)\to X$. The image of this embedding (with the Riemannian metric induced from $X$) is a Riemannian submanifold of $X$ isometric to $(M,g)$. Now, if you wish, take $(M,g)$ to be say, the $d$-dimensional hyperbolic space (complete, simply-connected, constant curvature $-1$).

Edit. It is possible that you are using a nonstandard notion of a Riemannian submanifold and what you really mean is a totally geodesic submanifold. Then the only complete negatively curved totally geodesic submanifolds of $X$ (of dimension $\ge 2$) are hyperbolic planes.

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