In this section, we study the sectional curvature of \(\mathbb {H}\) equipped with a specific left invariant weak Riemannian metric. Following Sect. 3, we define the unique left invariant weak Riemannian metric \(\sigma \), such that
$$\begin{aligned} \sigma _0((v_1,v_2,v_3),(w_1,w_2,w_3))=g_0((v_1,v_2),(w_1,w_2))+v_3w_3 \end{aligned}$$
for \((v_1,v_2,v_3),(w_1,w_2,w_3)\in T_0\mathbb {H}\), according to (10). We recall the formula
$$\begin{aligned} g_0((v_1,v_2),(w_1,w_2)) =\eta (v_1,w_1)+\eta (v_2,w_2) =\langle Av_1,w_1\rangle + \langle Av_2,w_2\rangle \end{aligned}$$
and \(Ax=\sum _{k=1}^\infty x_k/k\), \(x=\sum _{k=1}^\infty x_ke_k\in \ell ^2\). For every positive integer j, we use the notation
$$\begin{aligned} e^1_j=(e_j,0,0),\quad e^2_j=(0,e_j,0)\quad \text {and} \quad e^3=(0,0,1), \end{aligned}$$
to indicate the standard orthonormal basis of \(\mathbb {H}\) seen as the Hilbert space \(\ell ^2\times \ell ^2\times {\mathbb {R}}\).
One easily realizes that the natural linear isomorphism between \({{\,\mathrm{Lie}\,}}(\mathbb {H})\) and \(\mathbb {H}\) is also an isomorphism of Lie algebras, where we equip \(\mathbb {H}\) with the Lie product (1). Thus, by slight abuse of notation, the left invariant vector fields of \({{\,\mathrm{Lie}\,}}(\mathbb {H})\) isomorphically associated with the basis \(e^1_j,e^2_j,e^3\) are denoted by the same symbols. It follows that for all \(i,j\ge 1\) and \(l=1,2\) we have
$$\begin{aligned}{}[e^1_i,e^2_j]=2\delta _{ij} e^3\quad \text {and}\quad [e^l_i,e^l_j]=0, \end{aligned}$$
(17)
where \(e^1_i,e^1_j,e^3\) are now understood as left invariant vector fields of \({{\,\mathrm{Lie}\,}}(\mathbb {H})\).
Now we consider a left invariant weak Riemannian metric \(\upsilon \) on \(\mathbb {H}\). The associated scalar product on \({{\,\mathrm{Lie}\,}}(\mathbb {H})\) is denoted by \(\left\langle \cdot ,\cdot \right\rangle _\upsilon \). We consider two orthonomal vectors \(X,Y\in {{\,\mathrm{Lie}\,}}(\mathbb {H})\) with respect to \(\left\langle \cdot ,\cdot \right\rangle _\upsilon \). By virtue of [15, Theorem 5], the sectional curvature \(K_\upsilon (X,Y)\) of the plane in \({{\,\mathrm{Lie}\,}}(\mathbb {H})\) spanned by X and Y can be obtained by the adjoint operator \({{\,\mathrm{ad}\,}}(Y)^\top (X)\), that we now introduce. We define \({{\,\mathrm{ad}\,}}(Y)(Z)=[Y,Z]\) and consider (in case it exists) the unique vector \({{\,\mathrm{ad}\,}}(Y)^\top (X)\in {{\,\mathrm{Lie}\,}}(\mathbb {H})\) that satisfies the equalities
$$\begin{aligned} \left\langle [Y,Z],X \right\rangle _\upsilon =\left\langle {{\,\mathrm{ad}\,}}(Y)(Z),X \right\rangle _\upsilon =\left\langle Z,{{\,\mathrm{ad}\,}}(Y)^\top (X) \right\rangle _\upsilon \end{aligned}$$
(18)
for every \(Z\in {{\,\mathrm{Lie}\,}}(\mathbb {H})\). Then we define
$$\begin{aligned} B_\upsilon (X,Y)={{\,\mathrm{ad}\,}}(Y)^\top (X) \in {{\,\mathrm{Lie}\,}}(\mathbb {H}). \end{aligned}$$
(19)
In the case \(\upsilon \) is a strong Riemannian metric, [1, Definition 5.2.12], the existence of \(B_\upsilon (X,Y)\) is always ensured, but not for any weak Riemannian metric. For instance, in Remark 4.1 below, we show the nonexistence of \(B_\sigma (X,Y)\) for a specific choice of X and Y, where \(\sigma \) is the weak Riemannian metric introduced at the beginning of this section.
From formula (53) of [15], we have
$$\begin{aligned} K_\upsilon (X,Y)=\left\langle \delta ,\delta \right\rangle _\upsilon +2\left\langle \alpha ,\beta \right\rangle _\upsilon -3\left\langle \alpha ,\alpha \right\rangle _\upsilon -4\left\langle B_X,B_Y \right\rangle _\upsilon , \end{aligned}$$
(20)
where we define
$$\begin{aligned} \delta&=\frac{1}{2}\left( B_\upsilon (X,Y)+B_\upsilon (Y,X) \right) ,\quad \beta =\frac{1}{2}\left( B_\upsilon (X,Y)-B_\upsilon (Y,X) \right) , \quad \nonumber \\ \alpha&=\frac{1}{2}[X,Y] \end{aligned}$$
(21)
$$\begin{aligned} B_X&=\frac{1}{2}B_\upsilon (X,X)\quad \text {and}\quad B_Y=\frac{1}{2}B_\upsilon (Y,Y). \end{aligned}$$
(22)
The proof of Theorem 1.3 follows from the application of (20) with respect to \(\sigma \) on suitable choices of planes. We denote by \(\left\langle \cdot ,\cdot \right\rangle _\sigma \) the scalar product in \({{\,\mathrm{Lie}\,}}(\mathbb {H})\) induced by the left invariant weak Riemannian metric \(\sigma \). The associated norm on \({{\,\mathrm{Lie}\,}}(\mathbb {H})\) is denoted by \(\Vert \cdot \Vert _\sigma \). We assume that for \(X,Y\in {{\,\mathrm{Lie}\,}}(\mathbb {H})\) the adjoint
$$\begin{aligned} B_\sigma (X,Y)={{\,\mathrm{ad}\,}}(Y)^\top (X) \end{aligned}$$
with respect to \(\sigma \) exists. In this case, its scalar product with a vector \(Z\in {{\,\mathrm{Lie}\,}}(\mathbb {H})\) is assigned by the following formula
$$\begin{aligned} \left\langle {{\,\mathrm{ad}\,}}(Y)^\top (X),Z \right\rangle _\sigma =\left\langle [Y,Z],X \right\rangle _\sigma = 2\beta (\pi (Y),\pi (Z)) x^3, \end{aligned}$$
(23)
as a consequence of (1), where \(\pi :\mathbb {H}\rightarrow \ell ^2\times \ell ^2\) is the canonical projection defined by
$$\begin{aligned} X=(\pi (X),x^3)=(\pi (X),0)+x^3e^3. \end{aligned}$$
We use the fixed orthonormal basis \(e^1_j,e^2_j,e^3\) of \(\mathbb {H}\) with respect to the standard Hilbert product of \(\ell ^2\times \ell ^2\times {\mathbb {R}}\), getting
$$\begin{aligned} {{\,\mathrm{ad}\,}}(Y)^\top (X)=\sum _{j=1}^\infty [{{\,\mathrm{ad}\,}}(Y)^\top (X)]^1_j e^1_j+ \sum _{j=1}^\infty [{{\,\mathrm{ad}\,}}(Y)^\top (X)]^2_j e^2_j+[{{\,\mathrm{ad}\,}}(Y)^\top (X)]^3e^3. \end{aligned}$$
Formula (23) yields
$$\begin{aligned}&\sum _{j=1}^\infty \frac{1}{j}[{{\,\mathrm{ad}\,}}(Y)^\top (X)]^1_j Z^1_j+ \sum _{j=1}^\infty \frac{1}{j}[{{\,\mathrm{ad}\,}}(Y)^\top (X)]^2_j Z^2_j+[{{\,\mathrm{ad}\,}}(Y)^\top (X)]^3Z^3\nonumber \\&\quad =2\beta (\pi (Y),\pi (Z))x^3 \end{aligned}$$
(24)
for arbitrary \(Z=Z^3e^3+\sum _{j=1}^\infty Z^1_je^1_j+Z^2_je^2_j\). In the case \(X=\pi (X)\), formula (24) shows the existence of \({{\,\mathrm{ad}\,}}(Y)^\top (\pi (X))\) and yields
$$\begin{aligned} B_\sigma (\pi (X),Y)={{\,\mathrm{ad}\,}}(Y)^\top (\pi (X))=0. \end{aligned}$$
(25)
In the case \(X=e^3\), again (24) for \(Z=e^1_j\) and \(Z=e^2_j\) respectively, gives
$$\begin{aligned}{}[{{\,\mathrm{ad}\,}}(Y)^\top (e^3)]^1_j=2j\beta (\pi (Y),e^1_j)\quad \text {and} \quad [{{\,\mathrm{ad}\,}}(Y)^\top (e^3)]^2_j=2j\beta (\pi (Y),e^2_j). \end{aligned}$$
(26)
For \(Z=e^3\), applying (24) we get
$$\begin{aligned}{}[{{\,\mathrm{ad}\,}}(Y)^\top (e^3)]^3=0. \end{aligned}$$
(27)
Assuming the existence of \({{\,\mathrm{ad}\,}}(Y)^\top (e^3)\), we have shown that
$$\begin{aligned} B_\sigma (e^3,Y)={{\,\mathrm{ad}\,}}(Y)^\top (e^3)=2\sum _{j=1}^\infty j\beta (\pi (Y),e^1_j)e^1_j +2 \sum _{j=1}^\infty j\beta (\pi (Y),e^2_j)e^2_j. \end{aligned}$$
Writing \(Y=Y^3e^3+\sum _{j=1}^\infty (Y^1_j e^1_j+Y^2_je^2_j)\), we finally get
$$\begin{aligned} B_\sigma (e^3,Y)=2\sum _{j=1}^\infty j(Y^1_je^2_j-Y^2_je^1_j). \end{aligned}$$
(28)
Then the assumption about the existence of \(B_\sigma (e^3,Y)\) corresponds to the convergence of its series. The next remark shows a choice of Y for which the series (28) does not converge.
Remark 4.1
If we consider the vector
$$\begin{aligned} W=\sum _{j=1}^\infty \frac{e^1_j}{j}\in {{\,\mathrm{Lie}\,}}(\mathbb {H}), \end{aligned}$$
(29)
then it is easy to check that the series (28) representing \(B_\sigma (e^3,W)\) does not converge. As a consequence, the adjoint \({{\,\mathrm{ad}\,}}(W)^\top (e^3)\) cannot be defined. In addition, Arnold’s formula (20) for the sectional curvature of the plane \({{\,\mathrm{span}\,}}\left\{ W,e^3 \right\} \) does not apply.
Proposition 4.2
We consider the orthonormal elements \(W_k,e^3\in {{\,\mathrm{Lie}\,}}(\mathbb {H})\) with \(k\ge 1\) and
$$\begin{aligned} W_k=\left( \sum _{j=1}^kj^{-3} \right) ^{-1/2}\sum _{j=1}^k\frac{e^1_j}{j}\in {{\,\mathrm{Lie}\,}}(\mathbb {H}). \end{aligned}$$
As the subspace \({{\,\mathrm{span}\,}}\left\{ W_k,e^3 \right\} \) converges to \({{\,\mathrm{span}\,}}\{W_\infty ,e^3\}\) for \(k\rightarrow \infty \), with
$$\begin{aligned} W_\infty =\left( \sum _{j=1}^\infty j^{-3} \right) ^{-1/2}\sum _{j=1}^\infty \frac{e^1_j}{j}\in {{\,\mathrm{Lie}\,}}(\mathbb {H}), \end{aligned}$$
(30)
it follows that
$$\begin{aligned} K_\sigma (W_k,e^3)\rightarrow +\infty . \end{aligned}$$
(31)
The convergence of \({{\,\mathrm{span}\,}}\left\{ W_k,e^3 \right\} \) to \({{\,\mathrm{span}\,}}\{W_\infty ,e^3\}\) is considered in the Grassmannian of the 2-dimensional planes contained in \({{\,\mathrm{Lie}\,}}(\mathbb {H})\).
Proof
First of all, the pointwise convergence of \(W_k\) to \(W_\infty \) implies the convergence of \({{\,\mathrm{span}\,}}\left\{ W_k,e^3 \right\} \) to \({{\,\mathrm{span}\,}}\{W_\infty ,e^3\}\). To compute \(K_\sigma (W_k,e^3)\), we first apply (25), getting
$$\begin{aligned} B_\sigma (W_k,e^3)={{\,\mathrm{ad}\,}}(e^3)^\top (W_k)=0 \end{aligned}$$
(32)
for all \(k\ge 1\). From (28), it follows that \(B_\sigma (e^3,e^1_j)=2je^2_j\), hence
$$\begin{aligned} B_\sigma \left( e^3,\frac{e^1_j}{j} \right) =2e^2_j. \end{aligned}$$
The bilinearity of \(B_\sigma (\cdot ,\cdot )\) yields
$$\begin{aligned} B_\sigma (e^3,W_k)=2\left( \sum _{j=1}^kj^{-3} \right) ^{-1/2}\sum _{j=1}^ke^2_j \end{aligned}$$
(33)
From (21), taking \(\delta =\left( B_\sigma (W_k,e^3)+B_\sigma (e^3,W_k) \right) /2\), we obtain
$$\begin{aligned} \left\langle \delta ,\delta \right\rangle _\sigma = \frac{1}{4}\Vert B_\sigma (e^3,W_k)\Vert _\sigma ^2 =\left( \sum _{j=1}^\infty j^{-3} \right) ^{-1} \Big \Vert \sum _{j=1}^ke^2_j\Big \Vert _\sigma ^2= \left( \sum _{j=1}^\infty j^{-3} \right) ^{-1}\sum _{j=1}^k j^{-1}\nonumber \\ \end{aligned}$$
(34)
From (21), (22), (25) and (28), we find
$$\begin{aligned} \alpha =\frac{1}{2}B_\sigma (W_k,W_k)=\frac{1}{2}B_\sigma (e^3,e^3)=0. \end{aligned}$$
(35)
Finally, by formula (20), we have proved that
$$\begin{aligned} K_\sigma (W_k,e^3)=\left\langle \delta ,\delta \right\rangle _\sigma = \left( \sum _{j=1}^\infty j^{-3} \right) ^{-1}\sum _{j=1}^k j^{-1}\rightarrow +\infty \end{aligned}$$
(36)
as \(k\rightarrow \infty \). This concludes the proof. \(\square \)
Proof of Theorem 1.3
Following the notation of the present section, we define
$$\begin{aligned} a_{1j}=\sqrt{j}e^1_j\quad \text {and} \quad a_{2j}=\sqrt{j}e^2_j \end{aligned}$$
of \({{\,\mathrm{Lie}\,}}(\mathbb {H})\), that are orthonormal with respect to \(\left\langle \cdot ,\cdot \right\rangle _\sigma \) and do not commute. To apply (20) for finding \(K_\sigma (a_{1j},a_{2j})\), we use (21) and (22). Due to (25), we get
$$\begin{aligned} B_\sigma (a_{1j},a_{2j})=B_\sigma (a_{2j},a_{1j})=0. \end{aligned}$$
As a result, we have
$$\begin{aligned} K_\sigma (a_{1j},a_{2j})=-3\left\langle \alpha ,\alpha \right\rangle _\sigma =-\frac{3}{4} \Vert [a_{1j},a_{2j}]\Vert _\sigma ^2=-3j^2. \end{aligned}$$
(37)
Now we wish to compute \(K_\sigma (a_{1j},e^3)\) and \(K_\sigma (a_{2j},e^3)\). We first apply (25) and (28), getting
$$\begin{aligned} B_\sigma (e^l_j,e^3)={{\,\mathrm{ad}\,}}(e^3)^\top (e^l_j)=0, \quad B_\sigma (e^3,e^1_j)=2je^2_j\quad \text {and}\quad B_\sigma (e^3,e^2_j)=-2je^1_j\nonumber \\ \end{aligned}$$
(38)
for all \(l=1,2\) and \(k\ge 1\). From (21), taking \(\delta =\left( B_\sigma (a_{1j},e^3)+B_\sigma (e^3,a_{1j}) \right) /2\), we obtain
$$\begin{aligned} \left\langle \delta ,\delta \right\rangle _\sigma&=\frac{1}{4}\Vert B_\sigma (a_{1j},e^3)+B_\sigma (e^3,a_{1j})\Vert _\sigma ^2=\frac{1}{4}\Vert \sqrt{j}B_\sigma (e^3,e^1_j)\Vert _\sigma ^2= \frac{j}{4}\Vert 2je^2_j\Vert _\sigma ^2 \end{aligned}$$
(39)
$$\begin{aligned}&=j^3\left\langle e^2_j,e^2_j \right\rangle _\sigma =j^3\left\langle Ae^2_j,e^2_j \right\rangle =j^2. \end{aligned}$$
(40)
From (21), (22), (25) and (28), we find
$$\begin{aligned} \alpha =\frac{1}{2}B_\sigma (e^1_j,e^1_j)=\frac{1}{2}B_\sigma (e^3,e^3)=0. \end{aligned}$$
(41)
Due to the formula for the sectional curvature (20), we have established that
$$\begin{aligned} K_\sigma (a_{1j},e^3)=\left\langle \delta ,\delta \right\rangle _\sigma =j^2. \end{aligned}$$
(42)
In analogous setting \(\delta =\left( B_\sigma (a_{2j},e^3)+B_\sigma (e^3,a_{2j}) \right) /2\), we obtain
$$\begin{aligned} \left\langle \delta ,\delta \right\rangle _\sigma =\frac{1}{4}\Vert B_\sigma (e^3,a_{2j})\Vert _\sigma ^2= \frac{j}{4}\Vert B_\sigma (e^3,e^2_j)\Vert _\sigma ^2=\frac{j}{4}\Vert 2je^1_j\Vert _\sigma ^2= j^3\Vert e^1_j\Vert _\sigma ^2=j^2.\nonumber \\ \end{aligned}$$
(43)
Again (21), (22), (25) and (28) imply that
$$\begin{aligned} \alpha =\frac{1}{2}B_\sigma (e^2_j,e^2_j)=\frac{1}{2}B_\sigma (e^3,e^3)=0. \end{aligned}$$
(44)
Due to (20), we have also proved that
$$\begin{aligned} K_\sigma (a_{2j},e^3)=\left\langle \delta ,\delta \right\rangle _\sigma =j^2. \end{aligned}$$
(45)
Taking into account (42) and (45), setting \(b=e^3\), we have completed the proof. \(\square \)
Remark 4.3
A direct verification shows that the computations of sectional curvature, to prove Theorem 1.3, could be also carried out extending the finite dimensional formula of [18, Lemma 1.1] for the countable structure coefficients of \({{\,\mathrm{Lie}\,}}(\mathbb {H})\). These coefficients are obtained from the orthonormal vectors \(\sqrt{j}e^1_j,\sqrt{j}e^2_j,e^3\) of \({{\,\mathrm{Lie}\,}}(\mathbb {H})\) with respect to \(\left\langle \cdot ,\cdot \right\rangle _\sigma \).
Following the notation of the this section, the sequence of curves whose length converges to zero in the proof of Theorem 3.3 can be written as
$$\begin{aligned} \gamma ^j(t)&=\frac{ct^2}{2} e^1_j-cte^2_j+\frac{c^2t^3}{6}e^3\in \mathbb {H}\quad \text {and}\\ \alpha ^j(t)&= c\bigg (\frac{1}{2}-\frac{t^2}{2}\bigg )e^1_j+c(t-1) e^2_j + c^2 \bigg (\frac{1}{6} + \frac{t^3}{6} - \frac{t^2}{2} + \frac{t}{2}\bigg )e^3\in \mathbb {H}. \end{aligned}$$
It is interesting to notice that all such curves are contained in the span of the planes
$$\begin{aligned} {{\,\mathrm{span}\,}}\{e^1_j,e^2_j\},\quad {{\,\mathrm{span}\,}}\{e^1_j,e^3\}\quad \text {and}\quad {{\,\mathrm{span}\,}}\{e^2_j,e^3\}. \end{aligned}$$
When these planes are seen in the Lie algebra, Theorem 1.3 shows that their sectional curvature blows-up, as the length of the curves converges to zero.