Implement geom_GeomPathInterpolate() so geom_link() from ggforce will work

I created a ggplot with gradient line segments between points using geom_link2 from the ggforce package and it does what I want (pictured), but when I pipe it through ggplotly(), I get this warning Warning: geom_GeomPathInterpolate() has yet to be implemented in plotly. and it prints without the lines.

library(ggplot2)
library(plotly)
library(ggforce)

# df
data <- structure(list( Percent = c(0.32, 0.23, 0.75, 0.25, 0.482, 0.421, 0.5114, 0.3423, 0.27, 0.4324, 0.347, 0.377, 0.26, 
0.375, 0.18604, 0.241378, 0.3095, 0.348837209, 0.33333, 0.1875, 0.2820, 0.65, 0.72, 0.75, 0.81, 0.87, 0.8244),  date = structure(c(18262, 18293, 18322, 18353, 18383, 18414, 18444, 18475, 18506, 18536, 
18567, 18597, 18628, 18659, 18687, 18718, 18748, 18779, 18809, 18840, 18871, 18901, 18932, 18962, 18993, 19024, 19052), class = "Date")), row.names = c(NA, 
-27L), class = c("tbl_df", "tbl", "data.frame"))

# Create ggplot
test <- data %>% 
    ggplot( aes ( x = date, y = Percent,
                colour= Percent, 
                group = 1)) +   # Note sure why, but I have to add this
    geom_link2(  ) +
  geom_point ( ) + 
  scale_y_continuous(limits = c(0,1)) + 
  scale_colour_gradient2(low = "red", mid = "yellow" , high = "green", 
                         midpoint= 0.70,
                         limits = c(0,1)) 

# Looks good
test

# Prints warning/without line
test %>% plotly::ggplotly( ) 

Comienza en la ruta de conversión de ggplotly() para el punto de entrada no compatible geom_GeomPathInterpolate() y reproduce la advertencia con el ejemplo de geom_link2 proporcionado. Sigue cómo se convierte el geom y, a continuación, verifica que la línea de gradiente se renderice sin la advertencia y conserve los puntos existentes.