{"id":6340,"date":"2026-02-10T17:05:55","date_gmt":"2026-02-10T17:05:55","guid":{"rendered":"https:\/\/sinatootoonian.com\/?p=6340"},"modified":"2026-02-24T14:47:02","modified_gmt":"2026-02-24T14:47:02","slug":"synchronization-with-bimodal-spines","status":"publish","type":"post","link":"https:\/\/sinatootoonian.com\/index.php\/2026\/02\/10\/synchronization-with-bimodal-spines\/","title":{"rendered":"Synchronization with Bimodal Spines"},"content":{"rendered":"\n

Two important aspects were missing in the model from the last post<\/a>:<\/p>\n\n\n\n

    \n
  1. Self-inhibition should always be present.<\/li>\n\n\n\n
  2. The magnitude\/release probability of self-inhibition should increase when the parent granule cell is excited.<\/li>\n<\/ol>\n\n\n\n

    In this post we’ll update the model to incorporate these effects.<\/p>\n\n\n\n

    The approach will be to assume that the baseline self-inhibition that an MC receives upon spiking, $\\delta$, can increase to $\\Delta > \\delta$ if its granule cell was recently activated.<\/p>\n\n\n\n

    The phase plane for such a model is simple. The resets are purely horizontal and vertical. Synchronization will occur if the periods of the two oscillators are the same. The total voltage unit 1 has to increase by to reach threshold is $1 + \\delta_1 + \\Delta_1$, and similarly for unit 2. <\/p>\n\n\n\n

    Therefore, equating periods, our synchrony condition is $$\\underbrace{{1 + \\delta_1 + \\Delta_1 \\over I_1}}_{T_1} = \\underbrace{{1 + \\delta_2 + \\Delta_2 \\over I_2}}_{T_2}.$$<\/p>\n\n\n\n

    It’s interesting to note that if the magnitude of the inhibition was proportional to the current drive into each cell, then the two periods would be approximately the same. That is, if the inhibition the MCs received from their activated GCs was matched to their drives, then the condition above would self-reinforcing: MCs would be synchronized and thereby maintain the activation of the GCs, and the GCs, receiving synchronized inputs, would maintain their high activity states, strong inhibition onto the MCs, keeping them synchronized.<\/p>\n\n\n\n

    It’s clear that synchrony requires inhibition to be set appropriately. Can that be learned, perhaps during an initial burst of activity in response to the odour?<\/p>\n\n\n\n

    Connection to inference<\/h2>\n\n\n\n

    Let’s return now to the task of the olfactory system and see if we can connect it to synchrony. The olfactory system receives an a set of glomerular activations, $\\II = [I_1, \\dots, I_M]$, and has to infer the odour $\\xx$ that caused them. The relevant quantity here is the posterior probability on odours given glomerular activations, $p(\\xx|\\II)$. <\/p>\n\n\n\n

    To report the most probable odour we need to maximize $p(\\xx|\\II)$, or (equivalently, it’s logarithm). The way<\/em> we’ll report it is in the activity of a granule cell\/piriform cortical neuron: we’ll assign one neuron to each component of $\\xx$, and set its activity to maximize the probability above. So, something like $$ \\tau {dx_i \\over dt} = {\\partial \\log p(\\xx|\\II) \\over \\partial x_i}.$$<\/p>\n\n\n\n

    To actually specify these dynamics, we need to specify the posterior probability above. It is proportional to the prior on odours $p(\\xx)$, and the likelihood $p(\\II|\\xx)$ of observing the activations $\\II$ given a particular odour $\\xx$: $$ p(\\xx|\\II) \\propto p(\\II|\\xx) p(\\xx).$$ <\/p>\n\n\n\n

    Focusing on the likelihood for the moment, it’s natural to assume that the glomerular responses will be conditionally independent given the odour identity. That is, $$ p(\\II|\\xx) = \\prod_{i = 1}^M p(I_i|\\xx).$$<\/p>\n\n\n\n

    To model the individual terms in the product, we note each glomerulus receive input from thousands of ORNs. Perhaps the simplest model for such activations is a Poisson process, with rate determined by the affinity $A_{ij}$ of the glomerulus for each component of the odourant. So, $$p(I_i|\\xx) = \\text{Poisson}\\left(k = I_i | \\lambda = \\sum_{j} A_{ij} x_j\\right), $$ where $\\lambda$ is the rate of the Poission process and $k$ is the number of observed events.<\/p>\n\n\n\n

    Since $ \\text{Poisson}(k|\\lambda) = {\\lambda^k e^{-\\lambda} \\over k!},$ $$\\quad \\log \\text{Poisson}(k|\\lambda) = k \\log \\lambda – \\lambda + \\text{constant in $\\lambda$}. $$ For our glomerular responses, this will be $$ \\log \\text{Poisson}(I_i | \\xx) = I_i \\log \\left(\\sum_j A_{ij} x_j\\right) – \\sum_j A_{ij} x_j.$$<\/p>\n\n\n\n

    The gradient of this with respect to $x_j$ is $$ {\\partial \\log \\text{Poisson}(I_i | \\xx) \\over \\partial x_j} = {I_i \\over \\sum_j A_{ij} x_j} A_{ij} – A_{ij}.$$ <\/p>\n\n\n\n

    The dynamics of granule cell $x_j$ will combine contributions from all glomeruli and end up something like \\begin{align*}\\tau {dx_j \\over dt} &= \\sum_i {\\partial \\log \\text{Poisson}(I_i | \\xx) \\over \\partial x_j} + \\dots\\\\ &= \\sum_i \\left({I_i \\over \\sum_j A_{ij} x_j} – 1\\right) A_{ij} + \\dots \\end{align*}<\/p>\n\n\n\n

    Now recall our synchrony expression for the period of a neuron receiving inhibition from a spine: $$ T_i = {1 + \\delta_i + \\Delta_i \\over I_i}.$$ We can invert it to get the firing frequency, $$ f_i = {1 \\over T_i} = {I_i \\over 1 + \\delta_i + \\Delta_i}.$$<\/p>\n\n\n\n

    Now the inhibition that neuron receives will be coming from multiple granule cells (via the corresponding spines), so $$f_i = {I_i \\over 1 + \\sum_j (\\delta_{ij} + \\Delta_{ij})}.$$<\/p>\n\n\n\n

    Then, if the strengths of the inhibition is set just right<\/em>, $$ \\delta_{ij} + \\Delta_{ij} = A_{ij} x_j,$$ the firing frequency becomes $$f_i = {I_i \\over 1 + \\sum_j A_{ij} x_j} \\approx {I_i \\over \\sum_j A_{ij} x_j} $$<\/p>\n\n\n\n

    We can then plug this into our inference dynamics and write \\begin{align*}\\tau {dx_j \\over dt} &= \\sum_i \\left({I_i \\over \\sum_j A_{ij} x_j} – 1\\right) A_{ij} + \\dots\\\\ &\\approx \\sum_i {I_i \\over \\sum_j A_{ij} x_j} A_{ij} + \\dots\\\\ &\\approx \\sum_i f_i A_{ij} + \\dots \\end{align*}<\/p>\n\n\n\n

    Summary<\/h2>\n\n\n\n

    We’ve made a number of assumptions, but the picture that emerges is quite natural:<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    Here the mitral cells fire at a rate determined by their excitation by the OSN inputs, and inhibition by the granule cells. A granule cell, representing an odour, integrates the firing rates of all mitral cells (one shown above) weighted by the affinity of the mitral cell’s parent glomerulus for the odour the granule cell represents.<\/p>\n\n\n\n

    But notice that there’s no role for synchrony. Each granule cell integrates the contributions of each mitral cell linearly. There are no nonlinear terms, e.g. $f_i f_j$, that would promote simultaneous activation of mitral cells. What matters is how many<\/em> spikes within some time window, not the pattern in which they arrive. This likely reflects the Poisson statistics of the input, which carry no additional useful statistics beyond their rates.<\/p>\n\n\n\n

    But what if the statistics of the input did <\/em>carry more useful information than just the rates? Real-life odours don’t just appear as step-functions producing constant activation. They arrive on plumes<\/em>, and would produce cofluctuations in all glomeruli responding to them. Inferring which odours are present would then require sensitivity to these cofluctuations, perhaps through synchrony…<\/p>\n\n\n\n

    $$\\blacksquare$$<\/p>\n\n\n\n

    <\/p>\n","protected":false},"excerpt":{"rendered":"

    I update the activity-dependent synchrony model to make spines bimodal, increasing their inhibition when their parent GC is active.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,148],"tags":[34,159,158,155],"class_list":["post-6340","post","type-post","status-publish","format-standard","hentry","category-blog","category-research","tag-granule-cell","tag-inhibition","tag-spines","tag-synchronization"],"_links":{"self":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/6340","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/comments?post=6340"}],"version-history":[{"count":57,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/6340\/revisions"}],"predecessor-version":[{"id":6645,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/posts\/6340\/revisions\/6645"}],"wp:attachment":[{"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/media?parent=6340"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/categories?post=6340"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sinatootoonian.com\/index.php\/wp-json\/wp\/v2\/tags?post=6340"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}